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Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given, 

sin x + sin 2x + sin 3x = 0 

To solve the equation we need to change its form so that we can equate the t-ratios individually. 

For this we will be applying transformation formulae. While applying the Transformation formula we need to select the terms wisely which we want to transform. 

As, 

sin x + sin 2x + sin 3x = 0 

∴ we will use sin x and sin 3x for transformation as after transformation it will give sin 2x term which can be taken common.

{∵ sin A + sin B = 2 sin \((\frac{A+B}2)\) cos \((\frac{A-B}2)\)

⇒ sin 2x + 2 sin \((\frac{3x+x}2)\) cos \((\frac{3x-x}2)\) = 0

⇒ 2sin 2x cos x + sin 2x = 0 

⇒ sin 2x ( 2cos x + 1) = 0 

∴ either, sin 2x = 0 or 2cos x + 1 = 0 

⇒ sin 2x = sin 0 or cos x = - 1/2 = cos (π-\(\frac{π}3\)) = cos \(\frac{2π}3\)

If sin x = sin y, implies x = nπ + (– 1)ny, where n ∈ Z. 

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

Comparing obtained equation with standard equation, 

we have: 

2x = nπ or x = 2mπ ± \(\frac{2π}3\)

∴ x = \(\frac{nπ}2\) or x = 2mπ  ± \(\frac{2π}3\)where m,n ϵ Z ..ans

Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) tan x + tan 2x + tan 3x = 0

Now let us simplify,

tan x + tan 2x + tan 3x = 0

tan x + tan 2x + tan (x + 2x) = 0

On using the formula,

tan (A + B) = [tan A + tan B]/[1 – tan A tan B]

Therefore,

tan x + tan 2x + [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

(tan x + tan 2x) (1 + 1/(1- tan x tan 2x)) = 0

(tan x + tan 2x) ([2 – tan x tan 2x]/[1 – tan x tan 2x]) = 0

Now,

(tan x + tan 2x) = 0 or ([2 – tan x tan 2x]/[1 – tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [2 – tan x tan 2x] = 0

tan x = tan (-2x) or tan x tan 2x = 2

x = nπ + (-2x) or tax x [2tan x/(1 – tan² x)] = 2 [Using, tan 2x = 2 tan x/1 - tan² x]

3x = nπ or 2 tan² x/(1 - tan² x) = 2

3x = nπ or 2 tan² x = 2(1 – tan² x)

3x = nπ or 2 tan² x = 2 – 2tan² x

3x = nπ or 4 tan² x = 2

x = nπ/3 or tan² x = 2/4

x = nπ/3 or tan² x = 1/2

x = nπ/3 or tan x = 1/√2

x = nπ/3 or x = tan α [let 1/√2 be ‘α’]

x = nπ/3 or x = mπ + α

∴ the general solution is

Thus, x = nπ/3 or mπ + α, where α = tan^-11/√2, m, n ∈ Z.

(ii) tan x + tan 2x = tan 3x

Now let us simplify,

tan x + tan 2x = tan 3x

tan x + tan 2x – tan 3x = 0

tan x + tan 2x – tan (x + 2x) = 0

On using the formula,

tan (A+B) = [tan A + tan B] / [1 – tan A tan B]

Therefore,

tan x + tan 2x – [[tan x + tan 2x]/[1- tan x tan 2x]] = 0

(tan x + tan 2x) (1 – 1/(1- tan x tan 2x)) = 0

(tan x + tan 2x) ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0

Now,

(tan x + tan 2x) = 0 or ([– tan x tan 2x] / [1 – tan x tan 2x]) = 0

(tan x + tan 2x) = 0 or [– tan x tan 2x] = 0

tan x = tan (-2x) or -tan x tan 2x = 0

tan x = tan (-2x) or 2tan² x / (1 – tan² x) = 0 [Using, tan 2x = 2 tan x / 1-tan² x]

x = nπ + (-2x) or x = mπ + 0

3x = nπ or x = mπ

x = nπ/3 or x = mπ

∴ the general solution is

x = nπ/3 or mπ, where m, n ∈ Z.

(iii) tan 3x + tan x = 2 tan 2x

Now let us simplify,

tan 3x + tan x = 2 tan 2x

tan 3x + tan x = tan 2x + tan 2x

Then, upon rearranging we get,

tan 3x – tan 2x = tan 2x – tan x

On using the formula,

tan (A - B) = [tan A – tan B]/[1 + tan A tan B]

Therefore,

[(tan 3x – tan 2x) (1 + tan 3x tan 2x)]/[1 + tan 3x tan 2x] = [(tan 2x - tan x) (1 + tan x tan 2x)]/[1 + tan 2x tan x]

tan (3x – 2x) (1 + tan 3x tan 2x) = tan (2x – x) (1 + tan x tan 2x)

tan x [1 + tan 3x tan 2x – 1 – tan 2x tan x] = 0

tan x tan 2x (tan 3x – tan x) = 0

Therefore,

tan x = 0 or tan 2x = 0 or (tan 3x – tan x) = 0

tan x = 0 or tan 2x = 0 or tan 3x = tan x

x = nπ or 2x = mπ or 3x = kπ + x

x = nπ or x = mπ/2 or 2x = kπ

x = nπ or x = mπ/2 or x = kπ/2

∴ the general solution is

Thus, x = nπ or mπ/2 or kπ/2, where, m, n, k ∈ Z.

sin = \(\frac{1}2\) or \(\frac{-1}2\)

sin x = sin a 

Here a = 30° or - 30° 

X = nπ +(-1)ⁿ a 

So, the values of x are 

nπ ± \(\frac{π}6\),n ∈ Z.

Option C

cos²x = (2 - √3 sin x)² 

1-sin²x = 4+3 sin²x - 4√3 sin x 

4 sin 2x - 4√3 sin x+3 = 0 

sin x = \(\frac{\sqrt3}2\) 

x = \(\frac{π}3\)

Option A

\(\frac{sin^22x}{cos^22x}\) = 1

sin²2x = cos²2x 

sin²2x = 1- Sin²2x 

2 sin²2x = 1 

sin2x = \(\frac{1}{\sqrt2}\)

sin 2x = sin45 

So

2x = nπ = \(\frac{π}4\)

x = \(\frac{nπ}2+\frac{π}8\)

We know if cos x = cos a 

Then 

x = 2nπ ± a 

here cos x = cos\((\frac{2π}3)\)

when n = 0, 

x = \(\frac{2π}3\)

when n = 1,

x = 2π - \(\frac{2π}3\) = \(\frac{4π}3\)

x =   \(\frac{2π}3\), \(\frac{4π}3\)

Option B

1 - sin²x + sin x+1 = 0 

sin²x - sin x - 2 = 0 

sin x = -1 

so, x= \(\frac{3π}2\)

and it lies between (\(\frac{5π}4\), \(\frac{7π}4\)) 

Option D

4sin x – 3cos x = 7 

4sin x – 7 = 3cos x 

Squaring both sides 

16(sin x)² + 49 – 56sin x = 9(cos x)² 

16(sin x)² + 49 – 56sin x = 9((sin x)²-1) 

16(sin x)² - 9(sin x)² - 56sin x + 49 + 9 = 0 

7(sin x)² - 56sin x + 58 = 0 

Solving the quadratic equation 

Sin x = 6.7774 or 1.2225 

But we know that sinθ lies between [-1,1] 

So there are no solutions for this given equation

\(\frac{\sqrt3}2\) sin x - \(\frac{cosx}2\) = a

cos30°sin x – sin30°cos x = a 

sin (x - 30) = a 

As the range of sin function is from [-1,1] 

So the value of a can be R-[-1,1] 

i.e. a ∈ (-∞, -2) ∪ (2, ∞)

As cos x = cos θ 

Then x = 2nπ ± θ 

And it is said that it has exactly one solution. 

So θ = 0 and 

x = \(\frac{2nπ}2\)

=nπ 

In the given interval taking n = 1,x = π {n = 0 is not possible as cos 0 = 1 not -1 but cos π is -1}

2y = 1 

i.e. y = \(\frac{1}2\)

and y = cos x 

so, to get the intersection points we must equate both the equations 

i.e. cos x = \(\frac{1}2\)

so, cos x = cos 60° 

and we know if cos x = cos a 

then x = 2nπ ± a where a ϵ [0, π] 

so here 

x = 2nπ ± \(\frac{π}3\)

So the possible values which belong [0,2π] are \(\frac{π}3\)and \(\frac{5π}3\). 

There are a total of 2 points of intersection.

\(\frac{sinx}{cosx}+\frac{1}{cosx}\) = 2 x cos x

sin x + 1 = 2 × (cos x)² 

sin x + 1 = 2 × (1 - (sin x)²) 

sin x + 1 = 2 – 2(sin x)² 

2(sin x)² + sin x - 1 = 0 

Consider a=sin x 

So, the equation will be 

2a²+a -1 = 0 

From the equation a=0.5 or -1 

Which implies 

Sin x=0.5 or sin x=(-1) 

Therefore x=30° or 270° 

But for x=270° our equation will not be defined as cos (270° )=0

So, the solution for x = 30° 

According to trigonometric equations 

If sin x=sin a 

Then x = nπ – na 

Here sin x = sin30 

So, x = nπ + (-1)ⁿ × 30 

For n = 0, x = 30 and n =1,x = 150° and for n = 2,x = 390 

Hence between 0 to 2π there are only 2 possible solutions.

Y = cosec x and y = - \(\frac{1}2\)

So

\(\frac{1}{sinx}\) = \(\frac{-1}2\)

Sin x = -2 

Which is not possible 

So 

There are 0 points of intersection.

Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) Given as sin² x – cos x = 1/4

Now let us simplify,

sin² x – cos x = 1/4

1 – cos² x – cos x = 1/4 [since, sin² x = 1 – cos² x]

4 – 4 cos² x – 4 cos x = 1

4cos² x + 4cos x – 3 = 0

Suppose cos x be ‘k’

Therefore,

4k² + 4k – 3 = 0

4k² – 2k + 6k – 3 = 0

2k(2k – 1) + 3(2k – 1) = 0

(2k – 1) + (2k + 3) = 0

(2k – 1) = 0 or (2k + 3) = 0

k = 1/2 or k = -3/2

cos x = 1/2 or cos x = -3/2

Then, we shall consider only cos x = 1/2. cos x = -3/2 is not possible.

Therefore,

cos x = cos 60° = cos π/3

x = 2nπ ± π/3

∴ the general solution is

Thus, x = 2nπ ± π/3, where n ϵ Z.

(ii) 2 cos² x – 5 cos x + 2 = 0

Now let us simplify,

2 cos² x – 5 cos x + 2 = 0

Suppose cos x be ‘k’

2k² – 5k + 2 = 0

2k² – 4k – k +2 = 0

2k(k – 2) - 1(k - 2) = 0

(k – 2) (2k – 1) = 0

k = 2 or k = 1/2

cos x = 2 or cos x = 1/2

Then, we shall consider only cos x = 1/2. cos x = 2 is not possible.

Therefore,

cos x = cos 60° = cos π/3

x = 2nπ ± π/3

∴ the general solution is

x = 2nπ ± π/3, where n ϵ Z.

(iii) Given as 2 sin² x + √3 cos x + 1 = 0

Now let us simplify,

2 sin² x + √3 cos x + 1 = 0

2 (1 – cos² x) + √3 cos x + 1 = 0 [since, sin² x = 1 – cos² x]

2 – 2 cos² x + √3 cos x + 1 = 0

2 cos² x – √3 cos x – 3 = 0

Suppose cos x be ‘k’

2k² – √3 k – 3 = 0

2k² - 2√3 k + √3 k – 3 = 0

2k(k – √3) + √3(k – √3) = 0

(2k + √3) (k – √3) = 0

k = √3 or k = -√3/2

cos x = √3 or cos x = -√3/2

Then, we shall consider only cos x = -√3/2. cos x = √3 is not possible.

Therefore,

cos x = -√3/2

cos x = cos 150° = cos 5π/6

Thus, x = 2nπ ± 5π/6, where n ϵ Z.

\(\frac{1}2\) cox + \(\frac{\sqrt3}2\) sinx = 1

cos 60 cos x + sin 60 sin x = 1 

cos (60 - x) = 1 

cos (60 - x) = cos 0° 

x = 60° 

Option D

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given, 

3sin² x – 5 sin x cos x + 8 cos² x = 2 

⇒ 3sin² x + 3 cos² x – 5sin x cos x + 5 cos² x = 2 

⇒ 3 - 5sin x cos x + 5 cos² x = 2 {∵ sin² x + cos²x = 1} 

⇒ 5cos² x + 1 = 5sin x cos x 

Squaring both sides: 

⇒ (5cos² x + 1)² = (5sin x cos x)² 

⇒ 25cos⁴ x + 10cos² x + 1 = 25 sin² x cos² x 

⇒ 25cos⁴ x + 10cos² x + 1 = 25 (1 - cos² x) cos² x 

⇒ 50cos⁴ x – 15 cos² x + 1 = 0 

⇒ 50cos⁴ x – 10 cos² x – 5cos² x + 1 = 0 

⇒ 10cos² x ( 5cos² x – 1) – (5cos² x – 1) = 0 

⇒ (10cos² x - 1)(5cos² x – 1) = 0 

∴ cos² x = 1/10 or cos² x = 1/5 

Hence, when cos² x = 1/10

We have, cos x = ± \(\frac{1}{\sqrt{10}}\)

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

let cos α = 1/√10 

∴ cos (π – α) = -1/√10 

∴ x = 2nπ ± α or x = 2nπ ± (π – α)

∴ when,  cos x = ± \(\frac{1}{\sqrt{10}}\)

x = 2nπ ± α or x = 2nπ ± (π – α) where n ∈ Z.  and cos α = \(\frac{1}{\sqrt{10}}\)

When cos² x = 1/5

We have, cos x = ± \(\frac{1}{\sqrt{5}}\).

If cos x = cos y, implies x = 2mπ ± y, where n ∈ Z. 

let cos β = 1/√5 

∴ cos (π – β) = -1/√5 

∴ x = 2mπ ± β or x = 2mπ ± (π – β)

∴ when,  cos x = ± \(\frac{1}{\sqrt{5}}\).

x = 2mπ ± β or x = 2mπ ± (π – β)  where n ∈ Z.  and cos β = \(\frac{1}{\sqrt{5}}\)...ans

Cos²x = (√3 -√3sin x)² 

1 - sin²x = 3 + 3sin²x - 6sin x 

4sin²x - 6sin x+2 = 0 

2sin²x - 3sin x+1 = 0 

sin x = 1 or 0.5 

We know, 

x = nπ + (-1)ⁿθ

x = nπ + (-1)ⁿ \((\frac{π}2)\) or x =  nπ + (-1)ⁿ \((\frac{π}6)\)

Therefore, the values of x are

\(\frac{π}6\),\(\frac{π}2\),\(\frac{5π}6\),π,\(\frac{13π}6\)

So, these values are obtained for different value of n from the equation

x =  nπ + (-1)ⁿ \(\frac{π}2\) - \(\frac{π}6\),n ∈ Z.

So, Option B

\(\sqrt3(\frac{1}{tanx}+tanx)\) = 4

\(\sqrt3(\frac{1+tan^2x}{tanx})\) = 4

√3+√3 tan²x = 4 tan x 

√3 tan²x - 4 tan x+√3 = 0 

Therefore

tan = √3 or tanx = \(\frac{1}{\sqrt3}\)

Therefore x = \(\frac{π}3\) or \(\frac{π}6\)

But here the smallest angle is \(\frac{π}6\)

Option C

1 - sin²x + sin x +1 = 0 

sin²x - sin x - 2 = 0 

sin x = - 1 

x = \(\frac{3π}2\)

Option D

Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) Given as 4 sin² x – 8 cos x + 1 = 0

Now let us simplify,

4 sin² x – 8 cos x + 1 = 0

4(1 – cos² x) – 8 cos x + 1 = 0 [since, sin² x = 1 – cos² x]

4 – 4 cos² x – 8 cos x + 1 = 0

4 cos² x + 8 cos x – 5 = 0

Suppose cos x be ‘k’

4k² + 8k – 5 = 0

4k² - 2k + 10k – 5 = 0

2k(2k – 1) + 5(2k – 1) = 0

(2k + 5) (2k – 1) = 0

k = -5/2 = -2.5 or k = 1/2

cos x = -2.5 or cos x = 1/2

Now, we shall consider only cos x = 1/2. cos x = -2.5 is not possible.

Therefore,

cos x = cos 60° = cos π/3

x = 2nπ ± π/3

∴ the general solution is

Thus, x = 2nπ ± π/3, where n ϵ Z.

(ii) tan² x + (1 – √3) tan x – √3 = 0

Now let us simplify,

tan² x + (1 – √3) tan x – √3 = 0

tan² x + tan x – √3 tan x – √3 = 0

tan x (tan x + 1) – √3 (tan x + 1) = 0

(tan x + 1) ( tan x – √3) = 0

tan x = -1 or tan x = √3

As, tan x ϵ (-∞ , ∞) therefore both values are valid and acceptable.

tan x = tan (-π/4) or tan x = tan (π/3)

x = mπ – π/4 or x = nπ + π/3

∴ the general solution is

x = mπ – π/4 or nπ + π/3, where m, n ϵ Z.

(iii) 3 cos² x – 2√3 sin x cos x – 3 sin² x = 0

Now let us simplify,

3 cos² x – 2√3 sin x cos x – 3 sin² x = 0

3 cos² x – 3√3 sin x cos x + √3 sin x cos x – 3 sin² x = 0

3 cos x (cos x – √3sin x) + √3 sin x (cos x – √3 sin x) = 0

√3 (cos x – √3 sin x) (√3 cos x + sin x) = 0

cos x – √3 sin x = 0 or sin x + √3 cos x = 0

cos x = √3 sin x or sin x = -√3 cos x

tan x = 1/√3 or tan x = -√3

As, tan x ϵ (-∞ , ∞) therefore both values are valid and acceptable.

tan x = tan (π/6) or tan x = tan (-π/3)

x = mπ + π/6 or x = nπ – π/3

∴ the general solution is

x = mπ + π/6 or nπ – π/3, where m, n ϵ Z.

(iv) Given as cos 4x = cos 2x

Now let us simplify,

cos 4x = cos 2x

4x = 2nπ ± 2x

Therefore,

4x = 2nπ + 2x [or] 4x = 2nπ – 2x

2x = 2nπ [or] 6x = 2nπ

x = nπ [or] x = nπ/3

∴ the general solution is

Thus, x = nπ [or] nπ/3, where n ϵ Z.

2sin²x = 3cos x 

2 - 2cos²x = 3cos x 

Solving the quadratic equation, we get 

cos x = \(\frac{1}2\)

Therefore x = 60° and 300° 

i.e.

θ  = \(\frac{π}3\) and \(\frac{5π}3\)

Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) Given as cos x + cos 2x + cos 3x = 0

Now let us simplify,

cos x + cos 2x + cos 3x = 0

Then, we shall rearrange and use transformation formula

cos 2x + (cos x + cos 3x) = 0

On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

cos 2x + 2 cos (3x + x)/2 cos (3x - x)/2 = 0

cos 2x + 2cos 2x cos x = 0

cos 2x (1 + 2 cos x) = 0

cos 2x = 0 or 1 + 2cos x = 0

cos 2x = cos 0 or cos x = -1/2

cos 2x = cos π/2 or cos x = cos (π – π/3)

cos 2x = cos π/2 or cos x = cos (2π/3)

2x = (2n + 1) π/2 or x = 2mπ ± 2π/3

x = (2n + 1) π/4 or x = 2mπ ± 2π/3

∴ the general solution is

x = (2n + 1) π/4 or 2mπ ± 2π/3, where m, n ϵ Z.

(ii) cos x + cos 3x – cos 2x = 0

Now let us simplify,

cos x + cos 3x – cos 2x = 0

Then, we shall rearrange and use transformation formula

cos x – cos 2x + cos 3x = 0

– cos 2x + (cos x + cos 3x) = 0

On using the formula, cos A + cos B = 2 cos (A + B)/2 cos (A - B)/2

– cos 2x + 2 cos (3x + x)/2 cos (3x - x)/2 = 0

– cos 2x + 2cos 2x cos x = 0

cos 2x (-1 + 2 cos x) = 0

cos 2x = 0 or -1 + 2cos x = 0

cos 2x = cos 0 or cos x = 1/2

cos 2x = cos π/2 or cos x = cos (π/3)

2x = (2n + 1) π/2 or x = 2mπ ± π/3

x = (2n + 1) π/4 or x = 2mπ ± π/3

∴ the general solution is

x = (2n + 1) π/4 or 2mπ ± π/3, where m, n ϵ Z.

(iii) sin x + sin 5x = sin 3x

Now let us simplify,

sin x + sin 5x = sin 3x

sin x + sin 5x – sin 3x = 0

Then, we shall rearrange and use transformation formula

– sin 3x + sin x + sin 5x = 0

– sin 3x + (sin 5x + sin x) = 0

On using the formula, sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2

– sin 3x + 2 sin (5x + x)/2 cos (5x - x)/2 = 0

2sin 3x cos 2x – sin 3x = 0

sin 3x ( 2cos 2x – 1) = 0

sin 3x = 0 or 2cos 2x – 1 = 0

sin 3x = sin 0 or cos 2x = 1/2

sin 3x = sin 0 or cos 2x = cos π/3

3x = nπ or 2x = 2mπ ± π/3

x = nπ/3 or x = mπ ± π/6

∴ the general solution is

x = nπ/3 or mπ ± π/6, where m, n ϵ Z.

(iv) Given as cos x cos 2x cos 3x = 1/4

Now let us simplify,

cos x cos 2x cos 3x = 1/4

4 cos x cos 2x cos 3x – 1 = 0

On using the formula,

2 cos A cos B = cos (A + B) + cos (A – B)

2(2cos x cos 3x) cos 2x – 1 = 0

2(cos 4x + cos 2x) cos2x – 1 = 0

2(2cos² 2x – 1 + cos 2x) cos 2x – 1 = 0 [using cos 2A = 2cos²A – 1]

4cos³ 2x – 2cos 2x + 2cos² 2x – 1 = 0

2cos² 2x (2cos 2x + 1) -1(2cos 2x + 1) = 0

(2cos² 2x – 1) (2 cos 2x + 1) = 0

Therefore,
2cos 2x + 1 = 0 or (2cos² 2x – 1) = 0

cos 2x = -1/2 or cos 4x = 0 [using cos 2θ = 2cos²θ – 1]

cos 2x = cos (π – π/3) or cos 4x = cos π/2

cos 2x = cos 2π/3 or cos 4x = cos π/2

2x = 2mπ ± 2π/3 or 4x = (2n + 1) π/2

x = mπ ± π/3 or x = (2n + 1) π/8

∴ the general solution is

Thus, x = mπ ± π/3 or (2n + 1) π/8, where m, n ϵ Z.

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as –

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

given, 

sec x cos 5x + 1 = 0, 0 < x < \(\frac{π}2\)

⇒ sec x cos 5x = -1 

⇒ cos 5x = - cos x 

∵ - cos x = cos (π – x) 

∴ cos 5x = cos (π – x) 

If cos x = cos y, implies 2nπ ± y, where n ∈ Z. 

∴ 5x = 2nπ ± (π – x) 

⇒ 5x = 2nπ + (π – x) or 5x = 2nπ – (π – x) 

⇒ 6x = (2n+1)π or 4x = (2n-1)π

∴ x = (2n + 1) \(\frac{π}6\) or x = (2n - 1) \(\frac{π}4\)where n ∈ Z. 

But, 0 < x < \(\frac{π}2\)

∴ x = \(\frac{π}6\)and x = \(\frac{π}4\)....ans

Since, the general solution of any trigonometric equation is given as

sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

tan x = tan y, implies x = nπ + y, where n ∈ Z.

(i) Given as sin 2x = √3/2

Now, let us simplify,

sin 2x = √3/2

= sin(π/3)

∴ the general solution is

2x = nπ + (-1)ⁿ π/3, where n ϵ Z.

Thus, x = nπ/2 + (-1)ⁿ π/6, where n ϵ Z.

(ii) cos 3x = 1/2

Now, let us simplify,

cos 3x = 1/2

= cos (π/3)

∴ the general solution is

3x = 2nπ ± π/3, where n ϵ Z.

x = 2nπ/3 ± π/9, where n ϵ Z.

(iii) sin 9x = sin x

Now, let us simplify,

Sin 9x – sin x = 0

By using transformation formula,

Sin A – sin B = 2 cos(A + B)/2 sin(A - B)/2

Therefore,

= 2 cos(9x + x)/2 sin(9x - x)/2

=> cos 5x sin 4x = 0

Cos 5x = 0 or sin 4x = 0

Then, let us verify both the expressions,

Cos 5x = 0

Cos 5x = cos π/2

5x = (2n + 1)π/2

x = (2n + 1)π/10, where n ϵ Z.

sin 4x = 0

sin 4x = sin 0

4x = nπ

x = nπ/4, where n ϵ Z.

∴ the general solution is

x = (2n + 1)π/10 or nπ/4, where n ϵ Z.

(iv) sin 2x = cos 3x

Now, let us simplify,

sin 2x = cos 3x

cos (π/2 – 2x) = cos 3x [since, sin A = cos (π/2 – A)]

π/2 – 2x = 2nπ ± 3x

π/2 – 2x = 2nπ + 3x [or] π/2 – 2x = 2nπ – 3x

5x = π/2 + 2nπ [or] x = 2nπ – π/2

5x = π/2 (1 + 4n) [or] x = π/2 (4n – 1)

x = π/10 (1 + 4n) [or] x = π/2 (4n – 1)

∴ the general solution is

x = π/10 (4n + 1) [or] x = π/2 (4n – 1), where n ϵ Z.

(v) tan x + cot 2x = 0

Now, let us simplify,

tan x = – cot 2x

tan x = – tan (π/2 – 2x) [since, cot A = tan (π/2 – A)]

tan x = tan (2x – π/2) [since, – tan A = tan -A]

x = nπ + 2x – π/2

x = nπ – π/2

∴ the general solution is

x = nπ – π/2, where n ϵ Z.

(vi) tan 3x = cot x

Now, let us simplify,

tan 3x = cot x

tan 3x = tan(π/2 – x) [since, cot A = tan(π/2 – A)]

3x = nπ + π/2 – x

4x = nπ + π/2

x = nπ/4 + π/8

∴ the general solution is

Thus, x = nπ/4 + π/8, where n ϵ Z.

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given,

sin x = tan x

⇒ sin x = \(\frac{sin\,x}{cos\,x}\)

⇒ sin x cos x = sin x

⇒ sin x (cos x - 1) = 0

either, 

sin x = 0 or cos x = 1 

⇒ sin x = sin 0 or cos x = cos 0 

We know that, 

If sin x = sin y, implies x = nπ + (– 1)n y, where n ∈ Z 

∵ sin x = sin 0 

∴ y = 0 

And hence,

x = nπ where n ∈ Z

Also, 

If cos x = cos y, implies x = 2mπ ±y, where m∈ Z 

∵ cos x = cos 0 

∴ y = 0 

Hence, x is given by

x = 2mπ where m ϵ Z 

∴ x = nπ or 2mπ ,where m,n ϵ Z …ans

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given,

tan x = - \(\frac{1}{\sqrt3}\)

We know that tan x and cot x have negative values in the 2^nd and 4^th quadrant. 

While giving solution, we always try to take the least value of y. 

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e. negative angle)

tan x = tan \((-\frac{π}6)\)

If tan x = tan y then x = nπ + y, where n ∈ Z. 

For above equation y = \(-\frac{π}6\)

∴ x = nπ +\((-\frac{π}6)\) ,where n ϵ Z 

Or x = nπ \(-\frac{π}6\) ,where n ϵ Z 

Thus, 

x gives the required general solution for the given trigonometric equation.

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given,

cosec x= -\(\sqrt2\)

We know that sin x, and cosec x have negative values in the 3^rd and 4^th quadrant. 

While giving a solution, we always try to take the least value of y 

The fourth quadrant will give the least magnitude of y as we are taking an angle in a clockwise sense (i.e., negative angle) 

- √2 = -cosec \((\frac{π}4)\) = cosec \((-\frac{π}4)\) { ∵ sin - θ = - sin θ}

∴ cosec x = cosec\((-\frac{π}4)\)

⇒ sim x = sin \((\frac{-π}4)\)

If sin x = sin y ,then x = nπ + (– 1)ⁿy , where n ∈ Z. 

For above equation y = \(-\frac{π}4\)

∴ x = nπ + (-1)ⁿ \((-\frac{π}4)\) ,where n ϵ Z

Or x = nπ + (-1)ⁿ⁺¹ \((\frac{π}4)\),where n ϵ Z 

Thus, 

x gives the required general solution for given trigonometric equation.

To Find: General solution. 

Given: 4sin x cos x + 2sin x + 2cos x + 1 = 0 

⇒ 2sin x (2cos x + 1) + 2cos x + 1 = 0 

So (2cos x + 1) (2sin x + 1) = 0

cos x -1/2 = cos(2π)3 or x = -1/2 = sin 7π/6

Formula used: cos θ = cos α

θ = 2nπ ± α or sin θ = sin α

⇒ θ = mπ + (-1)^m α where n, m ∈ l

x = 2nπ ± 2π/3 or x = mπ + (-1)^m. 7π/6 where n, m ∈ l

So the general solution is x = 2nπ ± 2π/3 or x = mπ + (-1)^m. 7π/6 where n, m ∈ l

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿ y, where n ∈ Z.

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

• tan x = tan y, implies x = nπ + y, where n ∈ Z.

given,

2 cos² x – 5 cos x + 2 =0 

As the equation is of 2^nd degree, so we need to solve a quadratic equation. 

First we will substitute trigonometric ratio with some variable k and we will solve for k 

Let, cos x = k 

∴ 2k² – 5k + 2 = 0 

⇒ 2k² – 4k – k +2 = 0 

⇒ 2k(k – 2) -1(k -2) = 0 

⇒ (k – 2)(2k - 1) = 0 

∴ k = 2 or k = 1/2

⇒ cos x = 2 {which is not possible} or cos x = 1/2 (acceptable) 

∴ cos x = 1/2 

⇒ cos x = cos 60° = cos π/3 

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

On comparing our equation with standard form, we have 

y = \(\fracπ{3}\)

∴ x = 2nπ ± \(\fracπ{3}\)where n ϵ Z ..ans

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given, 

cos x + cos 2x + cos 3x = 0 

To solve the equation we need to change its form so that we can equate the t-ratios individually. 

For this we will be applying transformation formulae. While applying the Transformation formula we need to select the terms wisely which we want to transform. 

As, 

cos x + cos 2x + cos 3x = 0 

∴ we will use cos x and cos 2x for transformation as after transformation it will give cos 2x term which can be taken common. 

∴ cos x + cos 2x + cos 3x = 0 

⇒ cos 2x + (cos x + cos 3x) = 0

{∵ cos A + cos B = 2 cos \((\frac{A+B}2)\) cos \((\frac{A-B}2)\)]

⇒ cos 2x + 2 cos \((\frac{3x+x}2)\) cos \(\frac{3x-x}2\) = 0

⇒ cos 2x + 2cos 2x cos x = 0 

⇒ cos 2x ( 1 + 2 cos x) = 0 

∴ cos 2x = 0 or 1 + 2cos x = 0 

⇒ cos 2x = cos π/2 or cos x = -1/2 

⇒ cos 2x = cos π/2 or cos x = cos (π - π/3) = cos (2π /3) 

If cos x = cos y implies x = 2nπ ± y, where n ∈ Z. 

From above expression and on comparison with standard equation we have: 

y = \(\frac{π}2\) or y = \(\frac{2π}3\)

∴ 2x = 2nπ ± \(\frac{π}2\) or x = 2mπ ± \(\frac{2π}3\)

∴ x = nπ ± \(\frac{π}4\) or x = 2mπ ± \(\frac{2π}3\) where m, n ϵ Z

Ideas required to solve the problem: 

The general solution of any trigonometric equation is given as – 

• sin x = sin y, implies x = nπ + (– 1)ⁿy, where n ∈ Z. 

• cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

• tan x = tan y, implies x = nπ + y, where n ∈ Z. 

Given,

cos 4x = cos 2x

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

From above expression and on comparison with standard equation we have: 

y = 2x

∴ 4x = 2nπ ± 2x

Hence,

4x = 2nπ + 2x or 4x = 2mπ - 2x

∴ 2x = 2nπ or 6x = 2mπ

⇒ x = nπ or x = \(\frac{2mπ}6\) = \(\frac{mπ}3\)

∴ x = nπ or \(\frac{mπ}3\)where m, n ϵ Z ..ans

To Find: General solution.

Given: tan³x - 3tanx = 0

⇒ tan x (tan² x - 3) = 0

⇒ tan x = 0 or tan x = ± √3

⇒ tan x = 0 or tan x = (π/3) or tan x = tan (2π/3)

Formula used: tan θ = 0 

⇒ θ = nπ, n ∈ l, tan θ = tan α

⇒ θ = kπ ± α, k ∈ l

So x = nπ or x = kπ + π/3 or x pπ + 2π/3 where n, k, p ∈ l

So general solution is x = nπ or x = kπ + π/3 or x = pπ + 2π/3 where n, k, p ∈ l