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Verified Answer :

\(2\cos\theta*\tan\theta + \tan\theta = 0 \\ \\2\cos\theta* \frac{\sin\theta}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = 0 \\ \\ 2\sin\theta + \frac{\sin\theta}{\cos\theta} = 0 \\ \\ \frac{2\sin{\theta}\cos{\theta}}{\cos{\theta}} + \frac{\sin\theta}{\cos\theta} = 0 \\ \\ \frac{2\sin{\theta}\cos{\theta}+\sin{\theta}}{\cos{\theta}} =0 \\ \\ 2\sin{\theta}\cos{\theta}+\sin{\theta}=0 \\ \\ \sin\theta(2\cos\theta+1)=0 \\ \\ \sin\theta=0 \Rightarrow \theta=0, \pi,2\pi \\ \\2\cos\theta=-1 \\ \\\cos\theta=- \frac{1}{2} \Rightarrow \theta= \frac{2\pi}{3} , \frac{4\pi}{3}\)


Therefore, \(\theta=0, \pi,2\pi ,\frac{2\pi}{3} , \frac{4\pi}{3}\)

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