Solve $\sin(3x)\sin(4x)=\cos(4x)\cos(5x)$ (problem from a Swedish 12th grade ’Student Exam’ from 1931)

Hint

Use https://mathworld.wolfram.com/WernerFormulas.html

$$\cos x-\cos7x=\cos x+\cos9x$$

Use $$\cos9x=-\cos7x=\cos(\pi-7x)$$

Can you take it from here?

Or use https://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$0=\cos9x+\cos7x=2\cos x\cos8x$$

$$\sin (3 x) \sin (4 x)=\frac{1}{2} (\cos (x)-\cos (7 x))$$ and $$\cos (4 x) \cos (5 x)=\frac{1}{2} (\cos (x)+\cos (9 x))$$ $$\frac{1}{2} (\cos (x)-\cos (7 x))=\frac{1}{2} (\cos (x)+\cos (9 x))$$ $$\cos(7x)+\cos(9x)=0$$ In the following equations $k$ is any integer $k\in\mathbb{Z}$ $$2\cos(8x)\cos x=0$$ $$\cos(8x)=0\to 8x=\pm\frac{\pi}{2}+2k\pi\to x=\pm\frac{\pi}{16}+\frac{k\pi}{4}$$ $$\cos x=0\to x=\frac{\pi}{2}+k\pi$$

Formulae can be found here.

Hint.

Calling $a = 4x$ and $b = x$ we have

$$ \sin(a-b)\sin (a)-\cos(a)\cos(a+b) = \cos(b)\sin^2(a)-\cos(b)\cos^2(a) = \cos(b)\left(\sin(a)-\cos(a)\right)\left(\sin(a)+\cos(a)\right)=0 $$

or

$$ \cos(x)(\sin(4x)-\cos(4x))(\sin(4x)+\cos(4x))=0 $$