Question

If $\cos 3x+\sin (2x-\frac{7\pi }{6})=-2$ then show that x is of the form $\frac{\pi }{3}(6m+1)$.

Answer 1

From the given relation we have
$1+\cos 3x+1-\cos [\frac{\pi }{2}+(2x-\frac{7\pi }{6})]=0$
$\because \sin \theta =-\cos (\frac{\pi }{2}+\theta )$
$2{\cos }^2\frac{3x}{2}+1-\cos (2x-\frac{2\pi }{3})=0$
or $2{\cos }^2\frac{3x}{2}+2{\sin }^2(x-\frac{\pi }{3})=0$
Above will hold when we have both
$\cos \frac{3x}{2}=0$ and $\sin (x-\frac{\pi }{3})=0$
$\frac{3x}{2}=\frac{\pi }{2},\frac{3\pi }{2}$ and $x-\frac{\pi }{3}=0,\pi ,2\pi ,$
$\therefore x=\frac{\pi }{3},\pi$ and $x=\frac{\pi }{3},\frac{4\pi }{3},\frac{7\pi }{3}$,...
$\therefore x=\frac{\pi }{3}$ is the common value which satisfies both.
$\therefore x=2n\pi +\frac{\pi }{3}=(6n+1)\frac{\pi }{3}$.