Question

$\sqrt{5+cosx}$= -√6 sinx

Answer 1

ОДЗ sinx≤0⇒x∈[3π/2+2πn;2π+2πn]
5+cosx=6sin²x
5+cosx-6+6cos²x=0
cosx=a
6a²+a-1=0
D=1+24=25
a1=(-1-5)/12=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn
x=4π/3+2πn
a2=(-1+5)/12=1/3⇒cosx=1/3⇒x=+-asrcos1/3+2πn
x=3π/2+arccos1/3+2πn