Question

The solution of the equation $(\sin x+\cos x{)}^{1+\sin 2x}=2,\pi \le x\le x\le \pi$ is

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is C

The maximum values of $\sin x+\cos x$ and
$1+\sin 2x$ are $\sqrt{2}$ and 2 respectively. Also, $\left(\sqrt{2}\right)2=2$
$\therefore$ The equation can hold only when $\sin x+\cos x=\sqrt{2}$ and $1+\sin 2x=2$
Now, $\sin x+\cos x=\sqrt{2}$
$\Rightarrow \cos (x-\frac{\pi }{4})=1\Rightarrow x=2n\pi +\frac{\pi }{4}$
$1+\sin 2x=2\Rightarrow \sin 2x=1$
$\Rightarrow x=\frac{n\pi }{2}+(-1{)}^n\frac{\pi }{4}$
The value in $[–\pi ,\pi ]$ satisfying both the equations is $\frac{\pi }{4}$ (when n = 0).