4 sin x + 2 = 0; [0, 2π)

Tammy Todd 2021-05-21 Answered
\(\displaystyle{4}{\sin{{x}}}+{2}={0};{\left[{0},{2}π\right)}\)

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jlo2niT
Answered 2021-05-22 Author has 19877 answers

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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content_user
Answered 2021-08-11 Author has 9769 answers

Answer is given below (on video)

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