Найдите корни уравнения sinx-√3cosx=1 на отрезке [-2π;4π]

Формула: [latex]a \sin x\pm b\cos x= \sqrt{a^2+b^2} \sin (x \pm \arcsin \frac{b}{ \sqrt{a^2+b^2} } )[/latex] [latex] \sqrt{a^2+b^2} = \sqrt{1+3}=2 [/latex] [latex]2\sin (x- \frac{\pi}{3} )=1 \\ \sin (x- \frac{\pi}{3})= \frac{1}{2} \\ x- \frac{\pi}{3}=(-1)^k\cdot \frac{\pi}{6}+ \pi k,k \in Z \\ x=(-1)^k\cdot \frac{\pi}{6}+\frac{\pi}{3}+\pi k,k \in Z[/latex] Отбор корней [latex]k=0;\,\,\,\, x= \frac{\pi}{2} [/latex] [latex]k=1;\,\,\,x= \frac{7 \pi }{6} [/latex] [latex]k=2;\,\,\, x= \frac{5 \pi}{2} [/latex] [latex]k=3;\,\,\,\, x= \frac{19 \pi}{6} [/latex] [latex]k=-1;\,\,\,\, x=- \frac{5 \pi }{6} [/latex] [latex]k=-2;\,\,\,\, x=- \frac{3 \pi }{2} [/latex]