Q. 3 B5.0( 1 Vote )

# Solve the followi

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 2nd 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 = �

cos x = 2 {which is not possible} or cos x = � (acceptable)

cos x = �

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 = π/3

where n ϵ Z ..ans

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