Question

The sum of all $x\in [0,\pi ]$ which satisfy the equation $sinx+\frac{1}{2}cosx=si{n}^2(x+\frac{\pi }{4})$ is.

• A. $\frac{\pi }{6}$
• B. $\frac{5\pi }{6}$
• C. $\pi$
• D. $2\pi$

Answer 1

Answer: (C)

$\pi$

$\sin x+\frac{\cos x}{2}={\sin }^2(x+\frac{\pi }{4})$

$\therefore \frac{2\sin x+\cos x}{2}={\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)}^2$

$\therefore \frac{2\sin x+\cos x}{2}=\frac{1+2\sin x\cos x}{2}$

$\therefore (2\sin x-1)(\cos x-1)=0$

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

$\therefore x=\frac{\pi }{6},\frac{5\pi }{6}$ or $x=0$

Sum of all $x=\frac{\pi }{6}+\frac{5\pi }{6}+0=\pi$