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Question Number 109489 by bobhans last updated on 24/Aug/20

$$\left(1\right)\sin \left(2x\right)-\cos \left(2x\right)-\sin \left(x\right)+\cos \left(x\right)=0$$$$\left(2\right)\underset{x\to 0}{\lim }\frac{e^x-e^{-x}}{\sin x}$$$$\left(3\right)\underset{x\to -1}{\lim }\mid x+1\mid \sin (x+1)$$$$$$

Answered by john santu last updated on 24/Aug/20

$$\left(2\right)\underset{x\to 0}{\lim }\frac{e^x+e^{-x}}{\cos x}=\frac{2}{1}=2$$$$\left(3\right)\underset{x\to -1}{\lim }\mid x+1\mid \sin (x+1)=0\times 0=0$$

Answered by john santu last updated on 24/Aug/20

$$\left(1\right)\sin \left(2x\right)-\sin \left(x\right)=\cos \left(2x\right)-\cos \left(x\right)$$$$\Rightarrow 2\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)=-2\sin \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)$$$$\Rightarrow 2\sin \left(\frac{x}{2}\right)\{\cos \left(\frac{3x}{2}\right)+\sin \left(\frac{3x}{2}\right)\}=0$$$$\{\begin{array}{l}2\sin \left(\frac{x}{2}\right)=0\Rightarrow \frac{x}{2}=2k\pi ;x=4k\pi \\ \tan \left(\frac{3x}{2}\right)=-1\Rightarrow \frac{3x}{2}=-\frac{\pi }{4}+k.\pi ;3x=-\frac{\pi }{2}+2k.\pi \end{array}$$$$\Rightarrow x=-\frac{\pi }{6}+\frac{2k}{3}.\pi$$

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