Solveeit Logo
Login

Question

Question: If \(6\cos 2\theta + 2\cos^{2}\frac{\theta}{2} + 2\sin^{2}\theta = 0, - \pi < \theta < \pi\), then \...

If 6cos⁡2θ+2cos⁡2θ2+2sin⁡2θ=0,−π<θ<π6\cos 2\theta + 2\cos^{2}\frac{\theta}{2} + 2\sin^{2}\theta = 0, - \pi < \theta < \pi, then θ\theta =

A

π3\frac{\pi}{3}

B

π3,cos⁡−1(35)\frac{\pi}{3},\cos^{- 1}\left( \frac{3}{5} \right)

C

cos⁡−1(35)\cos^{- 1}\left( \frac{3}{5} \right)

D

π3,π−cos⁡−1(35)\frac{\pi}{3},\pi - \cos^{- 1}\left( \frac{3}{5} \right)

Answer

π3,π−cos⁡−1(35)\frac{\pi}{3},\pi - \cos^{- 1}\left( \frac{3}{5} \right)

Explanation

Solution

The given equation can be written as

6(cos⁡2θ−1)+(1+cos⁡θ)+2(1−cos⁡2θ)=06\left( \cos^{2}\theta - 1 \right) + \left( 1 + \cos\theta \right) + 2\left( 1 - \cos^{2}\theta \right) = 0

⇒ 10cos⁡2θ+cos⁡θ−3=010\cos^{2}\theta + \cos\theta - 3 = 0

⇒(5cos⁡θ+3)(2cos⁡θ−1)=0\left( 5\cos\theta + 3 \right)\left( 2\cos\theta - 1 \right) = 0

⇒ cos⁡θ=12\cos\theta = \frac{1}{2} or cos⁡θ=−35\cos\theta = - \frac{3}{5}

⇒ θ=π3\theta = \frac{\pi}{3}or θ=π−cos⁡−1(35)\theta = \pi - \cos^{- 1}\left( \frac{3}{5} \right)as −π<θ<π- \pi < \theta < \pi

⇒ θ=π3,π−cos⁡−1(35)\theta = \frac{\pi}{3},\pi - \cos^{- 1}\left( \frac{3}{5} \right)