1. Zadatak

Primijenimo u danoj jednadžbi sljedeće identitete

$$\cos^2 x=\frac{1+\cos 2x}{2},\qquad \cos^2 2x=\frac{1+\cos 4x}{2}.$$

Slijedi

$$\frac{1+\cos 2x}{2}+\frac{1+\cos 4x}{2}+\cos^2 3x=1$$ $$1+\cos 2x+1+\cos 4x+2\cos^2 3x=2$$ $$\cos 2x+\cos 4x+2\cos^2 3x=0$$ $$2\cos\frac{2x+4x}{2}\cos\frac{2x-4x}{2}+2\cos^2 3x=0$$ $$2\cos 3x\cdot\cos x+2\cos^2 3x=0$$ $$2\cos 3x\cdot(\cos x+\cos 3x)=0$$ $$4\cos 3x\cdot\cos 2x\cdot\cos x=0.$$

Prema tome,

$$\cos 3x=0\implies x=\frac{\pi}{6}+\frac{k\pi}{3},\qquad k\in\mathbb Z,$$ $$\cos 2x=0\implies x=\frac{\pi}{4}+\frac{k\pi}{2},\qquad k\in\mathbb Z,$$ $$\cos x=0\implies x=\frac{\pi}{2}+k\pi,\qquad k\in\mathbb Z.$$