Question

√sin2x=√(cosx-sinx-1) [-7п/2;0]

Answer 1

sinx * sin2x=1-cos²x

sinx * 2sinx*cosx=sin²x

2 * sin²x* cosx-sin²x=0

sin²x(2cosx-1)=0

sin²x=0, sinx=0,  x=πn, n∈Z

2cosx-1=0,  cosx=1/2, x=±arccos1/2+2πk, x=±π/3+2πk, k∈Z