Solveeit Logo
Login

Question

Question: \(\frac{\cos 3\theta}{2\cos 2\theta - 1} = \frac{1}{2}\) if...

cos⁡3θ2cos⁡2θ−1=12\frac{\cos 3\theta}{2\cos 2\theta - 1} = \frac{1}{2} if

A

θ=nπ+π3\theta = n\pi + \frac{\pi}{3}

B

θ=2nπ±π3\theta = 2n\pi \pm \frac{\pi}{3}

C

θ=2nπ±π6\theta = 2n\pi \pm \frac{\pi}{6}

D

θ=2nπ+π6\theta = 2n\pi + \frac{\pi}{6}

Answer

θ=nπ+π3\theta = n\pi + \frac{\pi}{3}

Explanation

Solution

The given equality can be written as 4cos⁡3θ−3cos⁡θ2(2cos⁡2θ−1)−1=12\frac{4\cos^{3}\theta - 3\cos\theta}{2\left( 2\cos^{2}\theta - 1 \right) - 1} = \frac{1}{2}

⇒ (4cos⁡2θ−3)cos⁡θ4cos⁡2θ−3=12\frac{\left( 4\cos^{2}\theta - 3 \right)\cos\theta}{4\cos^{2}\theta - 3} = \frac{1}{2}

⇒ cos⁡θ=12\cos\theta = \frac{1}{2}

⇒ θ=2nπ±π3\theta = 2n\pi \pm \frac{\pi}{3}

[cos⁡θ≠±32, as for this value of θ L.H.S. of the given equation is not defined.]\left\lbrack \cos\theta \neq \pm \frac{\sqrt{3}}{2},\text{ as for this value of }\theta\text{ L.H.S. of the given equation is not defined.} \right\rbrack