2. Zadatak

Kako je $\cos3x=4\cos^3x-3\cos x$, $\sin3x=3\sin x-4\sin^3x$, dana jednadžba prelazi u $$\begin{aligned} \cos3x\cdot\frac{\cos3x+3\cos x}{4}-\sin3x\cdot\frac{3\sin x-\sin3x}{4}&=a,\\ \cos^23x+3\cos3x\cos x+\sin^23x-3\sin3x\sin x&=4a,\\ 1+3(\cos3x\cos x-\sin3x\sin x)&=4a,\\ 1+3\cos4x&=4a,\\ \cos4x&=\frac{4a-1}{3}. \end{aligned}$$ Da bi postojalo rješenje, mora biti zadovoljen uvjet $$-1\leq\frac{4a-1}{3}\leq1,$$ dakle dobivamo $-\frac12\leq a\leq1$.