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Question Number 130047 by liberty last updated on 22/Jan/21

$$\mathrm{Given}4\cos {}^2\mathrm{x}\sin \mathrm{x}-2\sin {}^2\mathrm{x}=3\sin \mathrm{x}$$$$\mathrm{where}-\frac{\pi }{2}⩽\mathrm{x}⩽\frac{\pi }{2}.\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}$$$$\mathrm{all}\mathrm{posibble}\mathrm{value}\mathrm{of}\mathrm{x}.$$

Answered by MJS_new last updated on 22/Jan/21

$$\sin x=s\Rightarrow {\cos }^{2}x=1-s^2$$$$(s^2+\frac{s}{2}-\frac{1}{4})s=0$$$$\Rightarrow s=0\vee s=-\frac{1}{4}\pm \frac{\sqrt{5}}{4}$$$$\Rightarrow x=0\vee x=-\frac{3\pi }{10}\vee x=\frac{\pi }{10}$$$$\Rightarrow \mathrm{answer}\mathrm{is}-\frac{\pi }{5}$$

Answered by Alepro3 last updated on 22/Jan/21

$$$$$$\cos {}^{2}x=1-\sin {}^{2}x\Rightarrow$$$$4\sin x-4\sin {}^{3}x-2\sin {}^{2}x=3\sin x\Rightarrow$$$$\sin x(4\sin {}^{2}x+2\sin x-\sin x)=0\Rightarrow$$$$onesolutionis\sin x=0\Rightarrow x=k\pi$$$$solvingthe{2}^{nd}degreeequationwehave$$$$\sin x_{1,2}=(-2\pm \sqrt{4+16})/8=(-1\pm \sqrt{5})/4$$$$so{x}_{1,2}={\sin }^{-1}\left[(-1\pm \sqrt{5})/4\right]+2k\pi and{x}_{3,4}=\pi -{\sin }^{-1}[\left[(1-\sqrt{5})/4\right]+2k\pi \Rightarrow$$$$thesumofallofthisvalueofxis$$$$\mathrm{\Sigma }x=\pi +5k\pi$$

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