Question #c20c8

1 Answer
Jun 6, 2017

cos^2x=1+sinx
sin^2x+cos^2x=1
Therefore, cos^2x=1-sin^2x
So,
1-sin^2x=1+sinx
sin^2x+sinx=0
sinx(sinx+1)=0
Use null factor law (keeping in mind interval (0, 2π))
sinx=0 and sinx+1=0

sinx=0
x=0,2pi

sinx+1=0
sinx=-1
x=(3pi)/2
Therefore, x=0,2pi,(3pi)/2