Question

solve the equation 2 cos^2 x = 3sinx, for 0^degree < x < 360^degree

Answer 1

Use \[
\cos^2x=1-\sin^2x
\]

Answer 2

\[
2 \cos^2 x = 3\sin x
\]Becomes \[
2 -2\sin^2 x = 3\sin x\]
Now let \(u=\sin x\)\[
2 -2u^2 = 3u
\]This is just a quadratic equation.

Solve for \(u\) and remember \(x=\sin^{-1}u\)

Answer 3

thanks! really helps me alot :)

Answer 4

Good luck!