Question

Evaluate:$\cos (\frac{3\pi }{2}+x)\cos (2\pi +x)[\cos (\frac{3\pi }{2}-x)+\cot (2\pi +x)]=1$

Answer 1

$\cos \left(\frac{3\pi }{2+x}\right)$ $cos(2\pi +x)$ $\left\{cos\right(\frac{3\pi }{2}-x)+cot(2x+x\left)\right\}$

$\Rightarrow -sinx.cosx\{-sinx+cotx\}$

$\Rightarrow \frac{-sinxcosx}{sinx}\{cosx-si{n}^{2}x\}=1$

$\Rightarrow \{-cosxcosx-2+co{s}^{2}x\}=1$

$\Rightarrow -co{s}^{2}x+1+co{s}^{3}x=1$

$\Rightarrow co{s}^2\{cos-1\}=0$

$cosx=0;cosx=1$

$\therefore x=\pi /2orx=0.$