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Question Number 140187 by mathsuji last updated on 05/May/21

$$Solutionequation:$$$$sin2x=1+\sqrt{2}cosx+cos2x$$

Answered by Ankushkumarparcha last updated on 05/May/21

$$Solution:\sin \left(2x\right)={2\cos }^2\left(x\right)+\sqrt{2}\cos \left(x\right)(\because \cos \left(2x\right)={2\cos }^2\left(x\right)-1)$$$$\sin \left(x\right)=\cos \left(x\right)+1/\sqrt{2}(\because \sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right))$$$$\sin (\frac{\mathrm{\Pi }}{4}-x)=1/2=>x=2n\mathrm{\Pi }+\frac{5\mathrm{\Pi }}{12},n\in \mathbb{Z}$$

Commented by mr W last updated on 05/May/21

$$or$$$$\cos x=0$$$$\Rightarrow x=n\pi +\frac{\pi }{2}$$

Commented by mathsuji last updated on 06/May/21

$$thankyousir$$

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