To solve the equation \(\cos x = -\sin^2 x - 1\) in the interval \([0, 2\pi)\), let's work through the equation step by step.
1. Rewrite the Equation:
[tex]\[
\cos x = -\sin^2 x - 1
\][/tex]
2. Use Pythagorean Identity:
Recall that \(\sin^2 x + \cos^2 x = 1\). Therefore, we can express \(\sin^2 x\) in terms of \(\cos x\):
[tex]\[
\sin^2 x = 1 - \cos^2 x
\][/tex]
Substitute \(\sin^2 x\) into the equation:
[tex]\[
\cos x = -(1 - \cos^2 x) - 1
\][/tex]
3. Simplify the Equation:
Simplify the right-hand side:
[tex]\[
\cos x = -1 + \cos^2 x - 1
\][/tex]
[tex]\[
\cos x = \cos^2 x - 2
\][/tex]
4. Rearrange the Equation:
Move all terms to one side to form a quadratic equation:
[tex]\[
\cos^2 x - \cos x - 2 = 0
\][/tex]
5. Solve the Quadratic Equation:
Let \(u = \cos x\). The equation becomes:
[tex]\[
u^2 - u - 2 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -1\), and \(c = -2\):
[tex]\[
u = \frac{1 \pm \sqrt{1 + 8}}{2}
\][/tex]
[tex]\[
u = \frac{1 \pm 3}{2}
\][/tex]
Thus, we have two solutions for \(u\):
[tex]\[
u = 2 \quad \text{and} \quad u = -1
\][/tex]
6. Evaluate the Cosine Values:
- \(\cos x = 2\) is not possible because the cosine function ranges between -1 and 1.
- \(\cos x = -1\) is valid.
7. Determine the Associated \(x\) Value:
If \(\cos x = -1\), the angle \(x\) in the interval \([0, 2\pi)\) is:
[tex]\[
x = \pi
\][/tex]
8. State the Final Solution:
The solution to the equation \(\cos x = -\sin^2 x - 1\) in the interval \([0, 2\pi)\) is:
[tex]\[
x = \pi
\][/tex]
Expressing the solution in terms of \(\pi\):
[tex]\[
x = \pi
\][/tex]
