# 4 sin x + 2 = 0; [0, 2π)

$$\displaystyle{4}{\sin{{x}}}+{2}={0};{\left[{0},{2}π\right)}$$

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jlo2niT

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:

$$x=7\frac{\pi}{6},11\frac{\pi}{6}$$

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Answer is given below (on video)