We are given: \(\displaystyle{4}{\sin{{x}}}+{2}={0}\)

Subtract 2 from both sides: \(\displaystyle{4}{\sin{{x}}}=−{2}\)

Divide both sides by 4: \(\displaystyle{\sin{{x}}}=−{12}\)

Recall that sine is negative in QIII and QIV.

The reference angle is \(x’=\frac{\pi}{6}\) since sin \(\frac{\pi}{6}=\frac{1}{2}\)

The QIII solution is:

\(x=\pi+x’=\pi+\frac{\pi}{6}=7\frac{\pi}{6}\)

The QIV solution is:

\(x=2\pi−x’=2\pi−\frac{\pi}{6}=11\frac{\pi}{6}\)

So, the solutions on \([0,2\pi)\) are:

\(x=7\frac{\pi}{6},11\frac{\pi}{6}\)