Question

If $\frac{-\pi }{2}\le x\le \frac{\pi }{2}$, then the two curves $y=\cos x$ and $y=\sin 3x$ intersect at

• A. $(\frac{\pi }{4},\frac{1}{\sqrt{2}})$ and $(\frac{\pi }{8},\cos \frac{\pi }{8})$
• B. $(-\frac{\pi }{4},\frac{1}{\sqrt{2}})$ and $(\frac{-\pi }{8},\cos \frac{\pi }{8})$
• C. $(\frac{\pi }{4},\frac{-1}{\sqrt{2}})$ and $(\frac{\pi }{8},-\cos \frac{\pi }{8})$
• D. $(-\pi /4,\pi /\sqrt{2})$

Answer 1

The correct option is A $(\frac{\pi }{4},\frac{1}{\sqrt{2}})$ and $(\frac{\pi }{8},\cos \frac{\pi }{8})$

These two curves intersect so at those points $\cos x=\sin 3x$

$\Rightarrow \sin 3x-\cos x=0\Rightarrow \sin 3x-\sin (\frac{\pi }{2}-x)=0$

$2\cos \left(\frac{3x+\frac{\pi }{2}-x}{2}\right)\sin \left(\frac{3x-\frac{\pi }{2}+x}{2}\right)=0$

$2\cos \left(\frac{2x+\frac{\pi }{2}}{2}\right)\sin \left(\frac{4x-\frac{\pi }{2}}{2}\right)=0\Rightarrow \cos (x+\frac{\pi }{4})\sin (2x-\frac{\pi }{4})=0$

$\Rightarrow \cos (x+\frac{\pi }{4})=0$ or $\sin (2x-\frac{\pi }{4})=0$

As $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

$\Rightarrow \cos (x+\frac{\pi }{4})=0\Rightarrow x+\frac{\pi }{4}=\frac{\pi }{2}\Rightarrow x=\frac{\pi }{4}\Rightarrow y=\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$

That is $(\frac{\pi }{4},\frac{1}{\sqrt{2}})$

$\Rightarrow \sin (2x-\frac{\pi }{4})=0\Rightarrow 2x-\frac{\pi }{4}=0\Rightarrow x=\frac{\pi }{8}\Rightarrow y=\cos \left(\frac{\pi }{8}\right)$

That is $(\frac{\pi }{8},\cos (\frac{\pi }{8}\left)\right)$

Hence , $\left(A\right)$