1/cos^2x=1-tanx1cos2x=1−tanx
1=cos^2x(1-tanx)1=cos2x(1−tanx)
1=cos^2x-tanxcos^2x1=cos2x−tanxcos2x
1=cos^2x-sinx/cosxcos^2x1=cos2x−sinxcosxcos2x
1=cos^2x-sinxcosx1=cos2x−sinxcosx
sin^2x+cos^2x=cos^2x-sinxcosxsin2x+cos2x=cos2x−sinxcosx
sin^2x=-sinxcosxsin2x=−sinxcosx
sin^2x+sinxcosx=0sin2x+sinxcosx=0
sinx(sinx+cosx)=0sinx(sinx+cosx)=0
sinx=0sinx=0 or sinx+cosx=0sinx+cosx=0
sinx=0sinx=0 or sinx=-cosxsinx=−cosx
Edit: from the comments:
For the range 0<=x<=2pi0≤x≤2π
a) sinx=0sinx=0
:.x=0, x=pi
b) sinx+cosx=0
We are going to divide both sides by cosx. It is necessary to check beforehand that we are not dividing by zero. However, if cosx=0, sinx=1, and the sum is not zero. In this case, we are safe to divide by cosx/
sinx/cosx+cosx/cosx=0
tanx+1=0
tanx=-1
x=(3pi)/4, (7pi)/4
Putting this together:
x=0, (3pi)/4, pi, (7pi)/4