## Task B-3.1.

The expression $\cos ^{2} x+\cos ^{2}\left(\frac{\pi}{6}+x\right)-2 \cos \frac{\pi}{6} \cdot \cos x \cdot \cos \left(\frac{\pi}{6}+x\right)$ has the same value for all real numbers $x$. Calculate that value.
## First Solution.

By factoring out the common factor, we get

$$
\begin{aligned}
& \cos ^{2} x+\cos ^{2}\left(\frac{\pi}{6}+x\right)-2 \cos \frac{\pi}{6} \cdot \cos x \cdot \cos \left(\frac{\pi}{6}+x\right)= \\
& =\cos ^{2} x+\cos \left(\frac{\pi}{6}+x\right)\left(\cos \left(\frac{\pi}{6}+x\right)-2 \cos \frac{\pi}{6} \cos x\right)
\end{aligned}
$$

By applying the cosine addition formula, we get

$$
\begin{array}{ll}
\cos ^{2} x+\left(\cos \frac{\pi}{6} \cos x-\sin \frac{\pi}{6} \sin x\right)\left(\cos \frac{\pi}{6} \cos x-\sin \frac{\pi}{6} \sin x-2 \cos \frac{\pi}{6} \cdot \cos x\right)= & \\
\cos ^{2} x-\left(\cos \frac{\pi}{6} \cos x-\sin \frac{\pi}{6} \sin x\right)\left(\cos \frac{\pi}{6} \cos x+\sin \frac{\pi}{6} \sin x\right) & 2 \text { points }
\end{array}
$$

Simplify the expression

$$
\cos ^{2} x-\left(\cos ^{2} \frac{\pi}{6} \cos ^{2} x-\sin ^{2} \frac{\pi}{6} \sin ^{2} x\right)
$$

Finally:

$$
\cos ^{2} x-\frac{3}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x=\frac{1}{4} \cos ^{2} x+\frac{1}{4} \sin ^{2} x=\frac{1}{4}
$$