Question

2sin(П/4 + х) * sin(П/4 - х) + sin^2x

Answer 1

5sin²x+4sin(π/2+x)=45sin²x+4cosx=45(1-cos²x)+4cosx=45-5cos²x+4cosx-4=01-5cos²x+4cosx=05cos²x-4cosx-1=0D=4²+1·5·4=16+20=36cosx=(4±√36)/10cosx=1 или cosx=-1/5x1=2πn, n∈Zx2=π-arccos0,2+2πnx3=arccos0,2-π+2πn