Question

Solve the following equation:
$\sin 3x-\cos 3x=\sqrt{\frac{3}{2}}$

Answer 1

$\frac{1}{\sqrt{2}}\sin 3x-\frac{1}{\sqrt{2}}\cos 3x=\frac{\sqrt{3}}{2}$

$\sin (3x-\frac{\pi }{4})=\frac{\sqrt{3}}{2}$

$3x-\frac{\pi }{4}=n\pi +{(-1)}^n\frac{\pi }{3}$

$x=\frac{n\pi }{3}+{(-1)}^n\frac{\pi }{9}+\frac{\pi }{12}.$