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Question Number 18092 by Tinkutara last updated on 15/Jul/17

$$\mathrm{A}\mathrm{value}\mathrm{of}\theta \mathrm{satisfying}$$$${4\cos }^2\theta \sin \theta -{2\sin }^2\theta =3\sin \theta \mathrm{is}$$$$\left(1\right)\frac{9\pi }{10}$$$$\left(2\right)\frac{\pi }{10}$$$$\left(3\right)-\frac{13\pi }{10}$$$$\left(4\right)-\frac{17\pi }{10}$$

Answered by ajfour last updated on 15/Jul/17

$$\left(\sin \theta \right)(4\cos {}^2\theta -2\sin \theta -3)=0$$$$\sin \theta =0,\mathrm{or}$$$$4-4\sin {}^2\theta -2\sin \theta -3=0$$$$\sin {}^2\theta +\frac{\sin \theta }{2}=\frac{1}{4}$$$$\Rightarrow {(\sin \theta +\frac{1}{4})}^2=\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$$$$\sin \theta =\frac{-1\pm \sqrt{5}}{4}$$$$\theta =\frac{\pi }{10},-\frac{3\pi }{10},\mathrm{and}\mathrm{so}\mathrm{on}..$$$$\mathrm{Among}\mathrm{the}\mathrm{options},\left(1\right)\mathrm{and}\left(2\right)\mathrm{with}$$$$\theta =\frac{\pi }{10},\frac{9\pi }{10}\mathrm{are}\mathrm{the}\mathrm{correct}\mathrm{options}.$$

Commented by Tinkutara last updated on 15/Jul/17

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

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