Решите уравнение (sin2x) ^4+1=cos3x

1) cos3x=√2:2; 3x=arccos√2/2=+-π/4+2πn; n∈z; x=π/12+2πn/3; n∈z

2) x/2+π/4=-arctg1=-π/4+πn; n∈z; x=π+2πn; n∈z

3) 2*sinx*cosx+cosx=0; cosx * (2sinx+1) = 0; cosx=0; x=π/2+πn; n∈z

2sinx+1=0; sinx=-0.5; x=-arcsin0,5=-π/6+2πn; n∈z; x=7π/6+2πn; n∈z

4) cos7x+cosx=2cos ((7x+x) / 2) cos ((7x-x) / 2) = 2cos4x*cos3x=0; cos4x=0; 4x=π/2+πn; n∈z; x=π/8+πn/4; n∈z; cos3x=0; 3x=π/2+πn; n∈z; x=π/6+πn/3; n∈z

5) 2cos²x=1+cos2x=2-2sin²x; 2 (1-sin²x+sinx) = 2,5; 1-sin²x+sinx=1.25; sinx-sin²x=0.25; sinx-sin²x-0.25=0; - sin²x+sinx+0.25=0; sinx=t; - 1≤t≤1; - t²+t+0.25=0; - 4t²+4t+1=0; D=32;

(-4+4√2) / - 8 = (1-√2) / 2; (-4-4√2) / - 8 = (1+√2) / 2-посторонний корень; sinx = (1-√2) / 2; x=arcsin ((1-√2) / 2) + 2πn; n∈z