Question

find all solutions to the equation.

2cos²x - 1 = 0

write your answer in radians in terms of π, and use the \or\ button as necessary.

example: x = π/5 + 2kπ, k ∈ z or x = π/7 +kπ, k ∈ z

Explanation:

Step1: Solve for \(\cos^{2}x\)

Given \(2\cos^{2}x - 1=0\), add \(1\) to both sides:
\(2\cos^{2}x=1\).
Then divide both sides by \(2\):
\(\cos^{2}x=\frac{1}{2}\).

Step2: Solve for \(\cos x\)

Take the square - root of both sides:
\(\cos x=\pm\frac{\sqrt{2}}{2}\).

Step3: Find the values of \(x\)

When \(\cos x = \frac{\sqrt{2}}{2}\), \(x = \frac{\pi}{4}+2k\pi,k\in\mathbb{Z}\) or \(x = 2\pi-\frac{\pi}{4}+2k\pi=\frac{7\pi}{4}+2k\pi,k\in\mathbb{Z}\).
When \(\cos x=-\frac{\sqrt{2}}{2}\), \(x=\pi-\frac{\pi}{4}+2k\pi=\frac{3\pi}{4}+2k\pi,k\in\mathbb{Z}\) or \(x=\pi+\frac{\pi}{4}+2k\pi=\frac{5\pi}{4}+2k\pi,k\in\mathbb{Z}\).
Combining these, we can write the general solution as \(x=\frac{\pi}{4}+k\pi,k\in\mathbb{Z}\) (since the period of \(\cos x\) is \(2\pi\) and the solutions \(\frac{\pi}{4}+2k\pi,\frac{3\pi}{4}+2k\pi,\frac{5\pi}{4}+2k\pi,\frac{7\pi}{4}+2k\pi\) can be grouped as \(\frac{\pi}{4}+k\pi\) where \(k\in\mathbb{Z}\)).

Answer:

\(x = \frac{\pi}{4}+k\pi,k\in\mathbb{Z}\)