Question

The solution of the equation ${[sinx+cosx]}^{1+sin2x}=2,-\pi \le x\le \pi$ is

• A.
• B. π
• C.
• D.

Answer 1

The correct option is C

$LetI={[sinx+cosx]}^{1+sin2x}={\left[\sqrt{2}sin(\frac{\pi }{4}+x)\right]}^{1+sin2x}$

$At,x=\frac{\pi }{4},$

$I={\left[\sqrt{2}sin(\frac{\pi }{4}+\frac{\pi }{4})\right]}^{1+sin\frac{2\pi }{4}}$

$={\left(\sqrt{2}\right)}^2=2$

Here, option (c) is correct.