Q. 3 A5.0( 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, 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

As, sin2 x = 1 – cos2 x

we have,   Let, cos x = k

4k2 + 4k – 3 = 0

4k2 -2k + 6k – 3

2k(2k – 1) +3(2k – 1) = 0

(2k – 1)(2k + 3) = 0

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

cos x = � or cos x = -3/2

As cos x lies between -1 and 1

cos x can’t be -3/2

So we ignore that value.

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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