The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi, 3 \pi]$ is :

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Trigonometric Equations

The number of solutions of the equation

cos (x + \frac{π}{3}) cos (\frac{π}{3} - x) = \frac{1}{4} cos^{2} 2 x, x \in [- 3 π, 3 π]

is :

Answer: (D)

Use the product-to-sum identity:

cos (A) cos (B) = \frac{1}{2} [cos (A + B) + cos (A - B)]

. Here

cos (\frac{π}{3} + x) cos (\frac{π}{3} - x) = cos^{2} (\frac{π}{3}) - sin^{2} x = \frac{1}{4} - sin^{2} x

. The equation becomes

4 (\frac{1}{4} - sin^{2} x) = cos^{2} 2 x

, which simplifies to

1 - 4 sin^{2} x = cos^{2} 2 x

. Using

sin^{2} x = \frac{1 - c o s 2 x}{2}

, we get

1 - 2 (1 - cos 2 x) = cos^{2} 2 x

, or

2 cos 2 x - 1 = cos^{2} 2 x

. Let

t = cos 2 x

t^{2} - 2 t + 1 = 0 \Rightarrow (t - 1)^{2} = 0 \Rightarrow t = 1

. Thus

cos 2 x = 1 \Rightarrow 2 x = 2 nπ \Rightarrow x = nπ

. In

[- 3 π, 3 π]

n \in {- 3, - 2, - 1, 0, 1, 2, 3}

, giving 7 solutions.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option