Matura maj 2024 (formuła 2023), zad. 10

$x \in \{ \frac{\pi}{6} , \frac{\pi}{4} , \frac{5 \pi}{6} , \frac{7 \pi}{6} , \frac{5 \pi}{4} , \frac{1 1 \pi}{6} \}$ .

Przekształcenia. $4 \cos ^{2} x - 3 = 2 ( 1 + \cos 2 x ) - 3 = 2 \cos 2 x - 1$ oraz $\sin 4 x = 2 \sin 2 x \cos 2 x$ . Równanie:

1) $\cos 2 x = \frac{1}{2}$ : $2 x = ± \frac{\pi}{3} + 2 k \pi$ , $x = ± \frac{\pi}{6} + k \pi$ . W $[0 , 2 \pi ]$ : $\frac{\pi}{6} , \frac{5 \pi}{6} , \frac{7 \pi}{6} , \frac{1 1 \pi}{6}$ .

2) $\sin 2 x = 1$ : $2 x = \frac{\pi}{2} + 2 k \pi$ , $x = \frac{\pi}{4} + k \pi$ . W $[0 , 2 \pi ]$ : $\frac{\pi}{4} , \frac{5 \pi}{4}$ . Odpowiedź: $x \in \{ \frac{\pi}{6} , \frac{\pi}{4} , \frac{5 \pi}{6} , \frac{7 \pi}{6} , \frac{5 \pi}{4} , \frac{1 1 \pi}{6} \}$ .