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Question Number 207561 by hardmath last updated on 18/May/24

$$4{\sin }^2\mathbf{x}+\sin 2\mathbf{x}=2$$$$\mathrm{find}:\mathbf{x}=?$$

Answered by mathzup last updated on 18/May/24

$$e\iff 4\frac{1-cos\left(2x\right)}{2}+sin\left(2x\right)=2\iff$$$$2-2cos\left(2x\right)+sin\left(2x\right)=2\iff$$$$sin\left(2x\right)-2cos\left(2x\right)=0\iff$$$$\sqrt{5}\{-\frac{2}{\sqrt{5}}cos\left(2x\right)+\frac{1}{\sqrt{5}}sin\left(2x\right)\}=0$$$$letcos\theta =-\frac{2}{\sqrt{5}}etsin\theta =\frac{1}{\sqrt{5}}\Rightarrow$$$$tan\theta =-\frac{1}{2}\Rightarrow \theta =-arctan\left(\frac{1}{2}\right)$$$$e\iff cos\left(2x\right)cos\theta +sin\left(2x\right)sin\theta =0$$$$\iff cos(2x-\theta )=0=cos(\frac{\pi }{2}+k\pi )$$$$\iff 2x-\theta =\frac{\pi }{2}+k\pi ou2x-\theta =-\frac{\pi }{2}+k\pi$$$$\iff 2x=\frac{\pi }{2}+\theta +k\pi ou2x=\theta -\frac{\pi }{2}+k\pi$$$$\iff x=\frac{\pi }{4}-\frac{1}{2}arctan\left(\frac{1}{2}\right)+\frac{k\pi }{2}ou$$$$x=-\frac{\pi }{4}+\frac{1}{2}arctan\left(\frac{1}{2}\right)+\frac{k\pi }{2}$$$$k\in Z$$

Commented by hardmath last updated on 19/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

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