Question

cos^2x+cosx-sin^2x=0

Answer 1

cos^2x+cosx-sin^2x=0
cos^2x+cosx-(1-cos^2x)=0
cos^2x+cosx-1+cos^2x=0
2cos^2x+cosx=0
cosx(2cosx+1)=0
cosx=0
x=П/2+Пn, n∈Z

2cosx+1=0
cosx=-1/2
x=±(П-arccos1/2)+2Пn, n∈Z
x=±2П/3+2Пn, n∈Z