Find the general solutions of the following equations : sin x = tan x

1.	Find the general solutions of the following equations : sin x = tan x
Answer» 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

Answer»

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

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