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

2*sin²x++sinx²=12*sin²x+sinx²=sin²x+cos²xsin²x-cos²x+sinx²=0-cos2x+sinx²=0sinx²-cos2x=0sinx²-sin(π/2-2x)=02*sin(x²+π/2-2x)/2•cos(x²-π/2+2x)/2=01)sin(x²+π/2-2x)/2=0(x²+π/2-2x)/2=πnx²+π/2-2x=2πnx²-2x+π/2-2πn=0D=1-π/2+2πn>0n>(π/2-1)/2πn>0,57/6,28x=(1±√D)/22)cos(x²-π/2+2x)/2=0x²+2x-π/2=π+2πnx²+2x-3π/2-2πn=0D1=1+3π/2+2πn>0n>-(1+3π/2)/2πx=(-1±√D1)/2