Find all solutions of the equation in the interval [tex][0, 2\pi)[/tex].

[tex]\cos x = -\sin^2 x - 1[/tex]

Write your answer in radians in terms of [tex]\pi[/tex].

If there is more than one solution, separate them with commas.



Answer :

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]

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