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Question: The solution of \(\sin 2\theta = \cos 3\theta\) is....

The solution of sin⁡2θ=cos⁡3θ\sin 2\theta = \cos 3\theta is.

A

⇒\Rightarrow

B

3θ=2nπ±(π2−2θ)3\theta = 2n\pi \pm \left( \frac{\pi}{2} - 2\theta \right)

C

⇒\Rightarrow

D

θ=2nπ5+π10\theta = \frac{2n\pi}{5} + \frac{\pi}{10}

Answer

⇒\Rightarrow

Explanation

Solution

θ=2nπ±π4\theta = 2n\pi \pm \frac{\pi}{4}

⇒θ=π42nπ±π4\theta = \frac{\pi}{4}2n\pi \pm \frac{\pi}{4}

tan⁡θ=cot⁡α⇒tan⁡θ=tan⁡(π2−α)\tan\theta = \cot\alpha \Rightarrow \tan\theta = \tan\left( \frac{\pi}{2} - \alpha \right) ⇒\Rightarrow

θ=nπ+π2−α3(sin⁡θ−cos⁡θ)=4sin⁡θcos⁡θ\theta = n\pi + \frac{\pi}{2} - \alpha 3(\sin\theta - \cos\theta) = 4\sin\theta\cos\theta 3(sin⁡θ−cos⁡θ)=2sin⁡2θ9(1−S)=4S2,3(\sin\theta - \cos\theta) = 2\sin 2\theta 9(1 - S) = 4S^{2},

S=sin⁡2θ4S2+9S−9=0S = \sin 2\theta 4S^{2} + 9S - 9 = 0.