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Question Number 17151 by Tinkutara last updated on 01/Jul/17

$$\mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${\cos }^2\theta -2\cos \theta =4\sin \theta -\sin 2\theta \mathrm{where}$$$$\theta \in [0,\pi ]\mathrm{is}$$

Answered by ajfour last updated on 01/Jul/17

$$4\sin \theta -2\sin \theta \cos \theta +2\cos \theta -\cos {}^2\theta =0$$$$2\sin \theta (2-\cos \theta )+\cos \theta (2-\cos \theta )=0$$$$(2-\cos \theta )(2\sin \theta +\cos \theta )=0$$$$\Rightarrow 2\sin \theta +\cos \theta =0$$$$\mathrm{or}\tan \theta =-\frac{1}{2}$$$$\mathrm{In}[0,\pi ],\theta =\pi -{\tan }^{-1}\left(\frac{1}{2}\right).$$

Commented by Tinkutara last updated on 01/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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