1/sin^2x-1/cos(3П/2-x)=2 ** [П; 5П/2]

Question

1/sin^2x-1/cos(3П/2-x)=2 на [П; 5П/2]

Comments

Тригонометрические уравнения помогите решить

Answer 1

1/sin²x+1/sinx=2
2sin²x-sinx-1=0
sinx=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒sinx=-1/2⇒x=(-1)^n+1*π/6+πn
a2=(1+3)/4=1⇒sinx=1⇒x=π/2+2πn

x=5π/2;7π/6;11π/6