Question

2sin^2x+sinx=1 solve for x

Answer 1

\[2x^2+x-1=0\\(x+1)(2x-1)=0\\x=-1,x=\frac{1}{2}\]
now replace \(x\) by \(\sin(x)\) and solve
\[\sin(x)=-1\] and \[\sin(x)=\frac{1}{2}\] for \(x\)

Answer 2

thank you!

Answer 3

yw