Question

8/ 1+tg^2x +2|sin x|=5.

Answer 1

1+tg²x=1/cos²x
8cos²x+2|sinx|-5=0
8-8sin²x+2|sinx|-5=0
8sin²x-2|sinx|-3=0
1)sinx<0⇒x∈(-π+2πn;2π+2πn,n∈z)
sinx=a
8a²+2a-3=0
D=4+96=100
a1=(-2-10)/16=-0,75⇒sinx=-0,75
x=π+arcsin0,75+2πn,n∈z
x=2π-arcsin0,75+2ππn,n∈Z
a2=(-2+10)/16=1/2⇒sinx=1/2не удов усл
2)sinx≥0⇒x∈[2πn,π+2πn,n∈Z]
8a²-2sinx-3=0
D=4+96=100
a1=(2-10)/16=-1/2⇒sinx=-1/2не удов усл
a2=(2+10)/16=0,75
x=arcsin0,75+2πn,n∈Z
x=π-arcsin0,75+2πn,n∈Z