Question

Find the point of intersection of the curves $y=cosx,y=sin3x$ if $-\pi /2\le x\le \pi /2$

• A. $(\frac{\pi }{8},\cos \frac{\pi }{8})$
• B. $(\frac{\pi }{4},\cos \frac{\pi }{4})$
• C. $(\frac{\pi }{2},\cos \frac{\pi }{2})$
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A $(\frac{\pi }{8},\cos \frac{\pi }{8})$
B $(\frac{\pi }{4},\cos \frac{\pi }{4})$

The point of intersection is given by
$\sin 3x=\cos x=\sin (\frac{\pi }{2}-x)\Rightarrow 3x=n\pi +{(-1)}^n(\frac{\pi }{2}-x)$
(1) let $n$ be even i.e $n=2m$
$\Rightarrow 3x=2m\pi +\frac{\pi }{2}-x\Rightarrow n=\frac{m\pi }{2}+\frac{\pi }{8}$
(2) let $n$ be odd i.e. $n=(2m+1)$
$\Rightarrow 3x=(2m+1)\pi -(\frac{\pi }{2}-x)$
$\Rightarrow 3x=2m\pi +\frac{\pi }{2}+x\Rightarrow x=m\pi +\frac{\pi }{4}$
Now as $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$
$\Rightarrow x=\frac{\pi }{8},\frac{\pi }{4},-\frac{3\pi }{8}$
Thus, point of intersection are
$(\frac{\pi }{8},\cos \frac{\pi }{8}),(\frac{\pi }{4},\cos \frac{\pi }{4})(-\frac{3\pi }{8},\cos \frac{3\pi }{8})$