Certainly! Let's solve the equation [tex]\(-\sin^2(x) = \cos(2x)\)[/tex] within the interval [tex]\([- \pi, \pi]\)[/tex].
### Step-by-Step Solution
1. Start with the given equation:
[tex]\[
-\sin^2(x) = \cos(2x)
\][/tex]
2. Simplify the equation:
Recall that the double-angle identity for cosine is:
[tex]\[
\cos(2x) = 1 - 2\sin^2(x)
\][/tex]
Substitute [tex]\(\cos(2x)\)[/tex] in the equation:
[tex]\[
-\sin^2(x) = 1 - 2\sin^2(x)
\][/tex]
3. Isolate the sine term:
Combine like terms:
[tex]\[
-\sin^2(x) + 2\sin^2(x) = 1
\][/tex]
[tex]\[
\sin^2(x) = 1
\][/tex]
4. Solve for [tex]\( \sin(x) \)[/tex]:
The possible values for [tex]\(\sin(x)\)[/tex] that satisfy [tex]\(\sin^2(x) = 1\)[/tex] are:
[tex]\[
\sin(x) = \pm 1
\][/tex]
5. Determine the corresponding [tex]\(x\)[/tex] values:
- For [tex]\(\sin(x) = 1\)[/tex], we have:
[tex]\[
x = \frac{\pi}{2} + 2k\pi
\][/tex]
- For [tex]\(\sin(x) = -1\)[/tex], we have:
[tex]\[
x = -\frac{\pi}{2} + 2k\pi
\][/tex]
Here, [tex]\(k\)[/tex] is any integer. But we need solutions within the interval [tex]\([- \pi, \pi]\)[/tex].
6. Identify solutions within the interval [tex]\([- \pi, \pi]\)[/tex]:
- For [tex]\(\sin(x) = 1\)[/tex]:
[tex]\[
x = \frac{\pi}{2}
\][/tex]
- For [tex]\(\sin(x) = -1\)[/tex]:
[tex]\[
x = -\frac{\pi}{2}
\][/tex]
7. List the final solutions:
The solutions within the interval [tex]\([- \pi, \pi]\)[/tex] are:
[tex]\[
x = -\frac{\pi}{2}, \frac{\pi}{2}
\][/tex]
### Summary
The solutions to the equation [tex]\(-\sin^2(x) = \cos(2x)\)[/tex] within the interval [tex]\([- \pi, \pi]\)[/tex] are:
[tex]\[
x = -\frac{\pi}{2}, \frac{\pi}{2}
\][/tex]
In decimal form, this corresponds to:
[tex]\[
x \approx -1.5708, 1.5708
\][/tex]
