Question
2sin(x)^2 = 2 + cos(x)
Answers
GPT-4o mini
To solve the equation \(2\sin^2(x) = 2 + \cos(x)\), we can first use the Pythagorean identity \(\sin^2(x) = 1 - \cos^2(x)\) to rewrite the equation in terms of \(\cos(x)\).
Starting with the original equation:
\[
2\sin^2(x) = 2 + \cos(x)
\]
Substituting \(\sin^2(x)\):
\[
2(1 - \cos^2(x)) = 2 + \cos(x)
\]
Expanding the left side:
\[
2 - 2\cos^2(x) = 2 + \cos(x)
\]
Next, we can rearrange the equation:
\[
-2\cos^2(x) - \cos(x) + 2 - 2 = 0
\]
This simplifies to:
\[
-2\cos^2(x) - \cos(x) = 0
\]
Factoring out a \(-\cos(x)\):
\[
-\cos(x)(2\cos(x) + 1) = 0
\]
Setting each factor equal to zero:
1. \(-\cos(x) = 0\) gives us:
\[
\cos(x) = 0
\]
This leads to the solutions:
\[
x = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z}
\]
2. \(2\cos(x) + 1 = 0\) gives us:
\[
2\cos(x) = -1 \quad \Rightarrow \quad \cos(x) = -\frac{1}{2}
\]
This leads to the solutions:
\[
x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z}
\]
Combining these solutions, we have:
\[
x = \frac{\pi}{2} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z}.
\]
Starting with the original equation:
\[
2\sin^2(x) = 2 + \cos(x)
\]
Substituting \(\sin^2(x)\):
\[
2(1 - \cos^2(x)) = 2 + \cos(x)
\]
Expanding the left side:
\[
2 - 2\cos^2(x) = 2 + \cos(x)
\]
Next, we can rearrange the equation:
\[
-2\cos^2(x) - \cos(x) + 2 - 2 = 0
\]
This simplifies to:
\[
-2\cos^2(x) - \cos(x) = 0
\]
Factoring out a \(-\cos(x)\):
\[
-\cos(x)(2\cos(x) + 1) = 0
\]
Setting each factor equal to zero:
1. \(-\cos(x) = 0\) gives us:
\[
\cos(x) = 0
\]
This leads to the solutions:
\[
x = \frac{\pi}{2} + k\pi \quad \text{for } k \in \mathbb{Z}
\]
2. \(2\cos(x) + 1 = 0\) gives us:
\[
2\cos(x) = -1 \quad \Rightarrow \quad \cos(x) = -\frac{1}{2}
\]
This leads to the solutions:
\[
x = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z}
\]
Combining these solutions, we have:
\[
x = \frac{\pi}{2} + k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{4\pi}{3} + 2k\pi \quad \text{for } k \in \mathbb{Z}.
\]