Решить уравнение

2 * (sin (x)) ^2+sin (x^2) = 1

2*sin²x++sinx²=1

2*sin²x+sinx²=sin²x+cos²x

sin²x-cos²x+sinx²=0

-cos2x+sinx²=0

sinx²-cos2x=0

sinx²-sin (π/2-2x) = 0

2*sin (x²+π/2-2x) / 2•cos (x²-π/2+2x) / 2=0

1) sin (x²+π/2-2x) / 2=0

(x²+π/2-2x) / 2=πn

x²+π/2-2x=2πn

x²-2x+π/2-2πn=0

D=1-π/2+2πn>0

n> (π/2-1) / 2π

n>0,57/6,28

x = (1±√D) / 2

2) cos (x²-π/2+2x) / 2=0

x²+2x-π/2=π+2πn

x²+2x-3π/2-2πn=0

D1=1+3π/2+2πn>0

n> - (1+3π/2) / 2π

x = (-1±√D1) / 2