Помогите решить 13 задание ........Пожалуйста..... Подробно

[latex]6cos^2x+5sinx-2=0\\\\6(1-sin^2x)+5sinx-2=0\\\\-6sin^2x+5sinx+4=0\\\\6sin^2x-5sinx-4=0\\\\D=121,\; (sinx)_1=\frac{5-11}{12}=-\frac{1}{2}\; ,\; \; (sinx)_2=\frac{5+11}{12}=\frac{4}{3}[/latex] [latex]a)\; sinx=\frac{4}{3}\ \textgreater \ 1\; \; \to \; \; net\; \; reshenij\\\\b)\; \; sinx=-\frac{1}{2}\; \; \to \; \; x=(-1)^{n}\cdot (-\frac{\pi}{6})+\pi n=(-1)^{n+1}\cdot \frac{\pi}{6}+\pi n,\; n\in Z[/latex] [latex]x= \left [ {{x=-\frac{\pi}{6}+\pi n} \atop {x=-\frac{5\pi}{6}+\pi n}} \right. \\\\-\frac{5\pi}{2} \leq x \leq -\pi :x=-\frac{13\pi}{6},-\frac{7\pi}{6},-\frac{\pi}{6},-\frac{5\pi}{6},\frac{\pi}{6}.[/latex]