Question

sin^2x-sinx= cos^2x
solve for x
a)over the domain of 0_ 15 years ago

Answer 1

\[\sin^2x-sinx=1-\sin^2x\]
\[2\sin^2x-sinx-1=0\]
\[u=sinx\]
\[2u^2-u-1=0\]
\[2u^2-2u+u-1=0\]
\[2u(u-1)+(u-1)=0\]
\[(u-1)(2u+1)=0\]
\[u=sinx=1\]
\[u=sinx=\frac{-1}{2}\]

Answer 2

too fast o.O