Question

Sin^2x+sinx-2> =0 тригонометрическое неравенство (полное решение)

Answer 1

Sinx=t, t∈[-1,1]
$t^{2}+t-2 \geq 0 \\ t^{2}+t-2=0 \\ t_{1} =1, t_{2} =-2 \\ t \leq -2,t \geq 1 \\ \left \{ {{t \leq -2,t \geq 1} \atop {-1 \leq t \leq 1}} \right. \\ t=1 \\ sin(x)=1 \\ x= \frac{ \pi}{2} +2 \pi n,$
n∈Z