Question

Solve for x:

cos^2x+sin(2x)-sinx-1=-sin^2x

Answer 1

\[\sin^2x-sinx+\sin2x+\cos^2x-1=0\]
\[1-sinx+\sin2x-1=0\]
\[-sinx+\sin2x=0\]
\[2\sin \frac{x }{ 2 }\cos \frac{ 3x }{ 2 }=0\]
\[\sin \frac{x }{ 2 }\cos \frac{ 3x }{ 2 }=0\]
\[\sin \frac{x }{ 2 }=0\]
x/2=npi
x=2npi
\[\cos \frac{ 3x }{ 2 }=0\]
\[\frac{ 3 }{ 2}x=\frac{ (2m+1)\pi }{ 2 }\]
\[x=\frac{ (2m+1)\pi }{ 3 }\]