Question

find all solutions to the equation.
2\cos 4x + 1 = 0
write your answer in radians in terms of \pi, and use the \or\ button as necessary.
example: x = \frac{\pi}{5} + 2k\pi, k \in \mathbb{z} or x = \frac{\pi}{7} + k\pi, k \in \mathbb{z}
help with this notation
x =

Explanation:

Step1: Solve for \(\cos4x\)

$$
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$$

Step2: Find the general solutions for \(4x\)

We know that \(\cos\theta=-\frac{1}{2}\) when \(\theta=\frac{2\pi}{3}+2k\pi\) or \(\theta=\frac{4\pi}{3}+2k\pi\), \(k\in\mathbb{Z}\). So for \(\theta = 4x\), we have:
\(4x=\frac{2\pi}{3}+2k\pi\) or \(4x=\frac{4\pi}{3}+2k\pi\), \(k\in\mathbb{Z}\)

Step3: Solve for \(x\)

For \(4x=\frac{2\pi}{3}+2k\pi\), divide both sides by \(4\):

$$x=\frac{\frac{2\pi}{3}+2k\pi}{4}=\frac{\pi}{6}+\frac{k\pi}{2}$$

For \(4x=\frac{4\pi}{3}+2k\pi\), divide both sides by \(4\):

$$x=\frac{\frac{4\pi}{3}+2k\pi}{4}=\frac{\pi}{3}+\frac{k\pi}{2}$$

Answer:

\(x = \frac{\pi}{6}+\frac{k\pi}{2},k\in\mathbb{Z}\) or \(x=\frac{\pi}{3}+\frac{k\pi}{2},k\in\mathbb{Z}\)