Question

2(sinx)^2-3cosx=0, why is solution 2*pi*n

Answer 1

\[\begin{align*}2\sin^2x-3\cos x&=0\\
2(1-\cos^2x)-3\cos x&=0\\
2\cos^2x+3\cos x-2&=0\\
(2\cos x-1)(\cos x+2)&=0
\end{align*}\]
Solve the following:
\[\begin{cases}2\cos x-1=0\\\cos x+2=0\end{cases}\]
You can eliminate the second equation, because there are no solutions to \(\cos x=-2\).

For the first equation, \(\cos x=\dfrac{1}{2}\), which gives \(\dfrac{\pi}{3}+2n\pi\) and \(-\dfrac{\pi}{3}+2n\pi\). Not sure how you're supposed to get just \(2n\pi\).

Answer 2

yeah there's a small chance the solution may be wrong...thanks!