Solve the trigonometric equation [calculate x]

Question

Calculate the x value: $\sin 3x \cdot \sin 5x= \sin 2x \cdot \sin 8x$

$x \neq 0$

I made $ \sin (a + b) $ until it was based on $ \cos x $ and $ \sin x $ and tried to solve the normal trigonometric equation, but couldn't or A solution for complex numbers? [Tchebyschev]

Answer 1

As said in a previous answer $$\sin (3x)\sin (5x)-\sin (2x) \sin(8x)=\sin ^2(x) (2 \cos (2 x)-2 \cos (6 x)-2 \cos (8 x)+1)$$ Now, for the part in brackets, let $t=\cos(2x)$ and expand to make $$2 \cos (2 x)-2 \cos (6 x)-2 \cos (8 x)+1=-16 t^4-8 t^3+16 t^2+8 t-1$$ which can be solved using radicals. The solutions are given as $$t_1=\frac{1}{8} \left(-1-\sqrt{5}-\sqrt{6 \left(5-\sqrt{5}\right)}\right)\qquad t_2=\frac{1}{8} \left(-1-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}\right)$$ $$t_3=\frac{1}{8} \left(-1+\sqrt{5}-\sqrt{6 \left(5+\sqrt{5}\right)}\right)\qquad t_4=\frac{1}{8} \left(-1+\sqrt{5}+\sqrt{6 \left(5+\sqrt{5}\right)}\right)$$ Back to the $x$'s $$x_1=\frac{13 \pi }{30}\qquad x_2=\frac{7 \pi }{30}\qquad x_3=\frac{11 \pi }{30}\qquad x_4=\frac{\pi }{30}$$ and remember the symmetry around $\frac \pi 2$.

Answer 2

Hint: Write your equation in the form $$\sin ^2(x) (-(-2 \cos (2 x)+2 \cos (6 x)+2 \cos (8 x)-1))=0$$