0\\(1-2sin^2x)+5sinx+2>0\\-2sin^2x+5sinx+3>0\\2sin^2x-5sinx-3<0\\sinx=t; |t|\leq 1\\2t^2-5t-3<0\\D=25+24=49\\t_{1.2}=\frac{5 \pm 7}{8}\\ t_1=3; t_2=-\frac{1}{2}\\ -\frac{1}{2}<t<3" alt="\displaystyle cos2x+5sinx+2>0\\(1-2sin^2x)+5sinx+2>0\\-2sin^2x+5sinx+3>0\\2sin^2x-5sinx-3<0\\sinx=t; |t|\leq 1\\2t^2-5t-3<0\\D=25+24=49\\t_{1.2}=\frac{5 \pm 7}{8}\\ t_1=3; t_2=-\frac{1}{2}\\ -\frac{1}{2}<t<3" align="absmiddle" class="latex-formula">
но |t|≤1. значит -1/2< sinx
-\frac{1}{2}\\ -\frac{\pi }{6}+2\pi n-\frac{1}{2}\\ -\frac{\pi }{6}+2\pi n