1
{sinx=0⇒x=πn,n∈z
{sin5x≠0⇒x≠πn/5
2
Возведем в квадрат
4+3cos3x-cos2x=6sin²x,sinx≥0⇒x∈[2πn;π+2πn]
4+3cosx-2cos²x+1-6+6cos²x=0
4cos²x+3cosx-1=0
cosx=a
4a²+3a-1=0
D=9+16=25
a1=(-3-5)/8=-1⇒cosx=-1⇒x=π+2πn
a2=(-3+5)/8=1/4⇒cosx=1/4⇒x=+-arccos1/4+2πn
Ответ x=π+2πn;x=arccos1/4+2πn,n∈z