Помогитееееее..........

$6sin^2x-sinx-1\leq 0\\D=1+4*6=5^2\\(sinx-\frac{1-5}{12})(x-\frac{1+5}{12})\leq 0;\\(sinx+\frac{1}{3})(sinx-\frac{1}{2})\leq 0;\\-1/3\leq sinx\leq 1/2\\\left[\begin{array}{ccc}-arcsin(1/3)+2\pi*n\leq x\leq arcsin(1/2)+2\pi*n\\\pi-arcsin(1/2)+2\pi*n\leq x\leq \pi+arcsin(1/3)+2\pi*n\\\end{array}$

Ответ: x∈[-arcsin(1/3)+2π*n;π/6+2π*n]∪[5π/6+2π*n;π+arcsin(1/3)+2π*n], n∈Z.