Question

The number of solutions of $\cos x=|1+\sin x|,0\le x\le 3\pi$, is

• A. 3
• B. 2
• C. 4
• D. none of these

Answer 1

Answer: (A)

3

given, $\cos x=\mid 1+\sin x\mid$

we know $-1\le sinx\le 1\Rightarrow 1+sinx\ge 0$

therefore,
$\cos x=1+\sin x\Rightarrow \cos x-\sin x=1$

$\Rightarrow \frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x=\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \left(\frac{\pi }{4}\right).\cos x-\sin \left(\frac{\pi }{4}\right).\sin x=\frac{1}{\sqrt{2}}$

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

Hence general solution is,
$x=2n\pi -\frac{\pi }{4}\pm \frac{\pi }{4}$

Thus solution in the given interval is,
$x=0,\frac{3\pi }{2},2\pi$
Hence, option 'A' is correct.