Question

The number of real solutions of the equation $\sqrt{1-cos2x}=\sqrt{2}si{n}^{-1}\left(sinx\right),-\pi <x\le \pi$, is:

• A. 1
• B. 2
• C. infinite

Answer 1

The correct option is B 2

$\sqrt{2}\sqrt{\left(si{n}^{2}x\right)}=\sqrt{2}si{n}^{-1}sinx,-\pi <x\le \pi$
$\left(i\right)0\le x\le \frac{\pi }{2},$ then $sinx=si{n}^{-1}sinx=x$.
X = 0 is obviously a solution and in $0<x\le \frac{\pi }{2},sinx<x$.
$\left(ii\right)\frac{\pi }{2}<x\le \pi$, we get
$sinx=si{n}^{-1}sin(\pi -x)=\pi -x$
$x=\pi$ is only solution in this case.
$\left(iii\right)-\frac{\pi }{2}\le x<0,-sinx=x$ which has no solution in this interval.
$\left(iv\right)-\pi <x<-\frac{\pi }{2}\Rightarrow -sinx=-\pi -x\Rightarrow sinx=\pi +x$
Which has no solution in the given interval.
Hence, there are only two solutions $x=0,\pi$