Question

The sum of solutions of the equation $\sin 2x-12(\sin x-\cos x)+12=0$ in the interval $(0,3\pi )$ is

• A. $\frac{3\pi }{2}$
• B. $4\pi$
• C. $7\pi$
• D. $\frac{5\pi }{2}$

Answer 1

The correct option is B $4\pi$

Given equation is
$\sin 2x-12(\sin x-\cos x)+12=0$
$\Rightarrow \sin 2x-12(\sin x-\cos x)+13=1$
$\Rightarrow -12(\sin x-\cos x)+13=1-\sin 2x$
$\Rightarrow -12(\sin x-\cos x)+13=(\sin x-\cos x{)}^2$
Let $t=\sin x-\cos x$
Then, $t^2+12t-13=0$
$\Rightarrow (t-1)(t+13)=0$
$\Rightarrow$ Either $t=1$ or $t=-13$ (Rejected)

When $t=1,$ we have
$\sin x-\cos x=1$
or, $\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}}\times 1$
or, $\sin (x-\frac{\pi }{4})=\sin \frac{\pi }{4}$
$\Rightarrow x-\frac{\pi }{4}=n\pi +(-1{)}^n\frac{\pi }{4}$
$\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{4}+\frac{\pi }{4},n\in Z$
As $x\in (0,3\pi ),$ $x=\frac{\pi }{2},\pi ,\frac{5\pi }{2}$