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Question: The general solution of \(\theta = 2k\pi\) is....

The general solution of θ=2kπ\theta = 2k\pi is.

A

θ=4kπ\theta = 4k\pi

B

k∈Ik \in I

C

tan⁡3θ−1tan⁡3θ+1=3\frac{\tan 3\theta - 1}{\tan 3\theta + 1} = \sqrt{3}

D

tan⁡3θ−tan⁡(π/4)1+tan⁡3θ.tan⁡(π/4)=3\frac{\tan 3\theta - \tan(\pi/4)}{1 + \tan 3\theta.\tan(\pi/4)} = \sqrt{3}

Answer

k∈Ik \in I

Explanation

Solution

θ=nπ±(π4−α2)\theta = n\pi \pm \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)

sin⁡6θ+sin⁡4θ+sin⁡2θ=0⇒2sin⁡4θcos⁡2θ+sin⁡4θ=0\sin 6\theta + \sin 4\theta + \sin 2\theta = 0 \Rightarrow 2\sin 4\theta\cos 2\theta + \sin 4\theta = 0 ⇒\Rightarrow.