Question

4cos^2(theta)+2sin(theta)-1=0

Answer 1

this is much easier than the last one
we can start by writing
\[4(1-\sin^2(\theta))+2\sin(\theta)-1=0\] and then solve the quadratic in \(\sin(\theta)\)

Answer 2

\[4-4\sin^2(\theta)+2\sin(\theta)-1=0\]
\[4\sin^2(\theta)-2\sin(\theta)-3=0\] etc