Solve the following equations : 2 cos2 x – 5 cos x + 2 = 0

1.	Solve the following equations : 2 cos2 x – 5 cos x + 2 = 0
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, 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

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,

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

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