Question

Помогите решить тригонометрию.

Answer 1

ОДЗ
sinx≤0⇒x∈[π+2πn;2π+2πn,n∈z]
(√2-√2сos²x+cosx-√2)*√(-6sinx)=0
-6sinx=0
sinx=0
x=πn,n∈z
2π≤πn≤7π/2
4≤2n≤7
2≤n≤3,5
n=2⇒x=2π
n=3⇒x=3π
√2cos²x-cosx=0
cosx(√2cosx-1)=0
cosx=0⇒x=π/2+πk,k∈z
x=5π/2
x=7π/2
cosx=1/√2
x=+-π/4+2πm,m∈z
x=9π/4
x={2π;3π;5π/2;7π/2;9π/4}