Sin3x=cos3x на отрезке [0;4]

Решите задачу:

$sin3x=cos3x\, |:cos3x\ne 0\\\\tg3x=1\\\\3x=\frac{\pi}{4}+\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{12}+\frac{\pi n}{3}\; ,\; n\in Z\\\\x\in [\, 0;4\, ]\; ,\; 4\; radiana\approx 4\cdot 57,3^\circ =229,2^\circ \; \in 3\; chetverti\; \to \\\\x_1=\frac{\pi}{12}\\\\x_2=\frac{\pi}{12}+\frac{\pi}{3}=\frac{5\pi}{12}\\\\x_3=\frac{\pi}{12}+\frac{2\pi}{3}=\frac{9\pi}{12}=\frac{3\pi}{4}$

$x_3=\frac{\pi}{12}+\frac{3\pi}{3}=\frac{13\pi}{12}$

Sin3x=cos3x ** отрезке [0;4]