Certainly! To solve the given trigonometric equation [tex]\(2 \cos^2 x + \sin x - 2 = 0\)[/tex], follow these steps:
1. Express [tex]\(\cos^2 x\)[/tex] in terms of [tex]\(\sin x\)[/tex]:
Using the Pythagorean identity, we know that:
[tex]\[
\cos^2 x = 1 - \sin^2 x
\][/tex]
2. Substitute [tex]\(\cos^2 x\)[/tex] into the equation:
Replacing [tex]\(\cos^2 x\)[/tex] with [tex]\(1 - \sin^2 x\)[/tex], the equation becomes:
[tex]\[
2(1 - \sin^2 x) + \sin x - 2 = 0
\][/tex]
3. Simplify the equation:
Distribute the 2 and combine like terms:
[tex]\[
2 - 2\sin^2 x + \sin x - 2 = 0
\][/tex]
[tex]\[
-2 \sin^2 x + \sin x = 0
\][/tex]
4. Factor the equation:
Factor out a [tex]\(\sin x\)[/tex] from the left-hand side:
[tex]\[
\sin x (-2 \sin x + 1) = 0
\][/tex]
5. Set each factor to zero and solve:
We have two factors to solve:
[tex]\[
\sin x = 0
\][/tex]
and
[tex]\[
-2 \sin x + 1 = 0
\][/tex]
6. Solve [tex]\(\sin x = 0\)[/tex]:
[tex]\[
x = 0, \pi, 2\pi, \ldots
\][/tex]
7. Solve [tex]\(-2 \sin x + 1 = 0\)[/tex]:
Rearrange the equation:
[tex]\[
\sin x = \frac{1}{2}
\][/tex]
The solutions for [tex]\(\sin x = \frac{1}{2}\)[/tex] within one period [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
x = \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
8. Combine all solutions:
The general solutions combining the periodic nature of the sine function are:
[tex]\[
x = 0, \pi, \frac{\pi}{6}, \frac{5\pi}{6}, 2\pi, \ldots
\][/tex]
Since we are asked for the primary solutions within one period [tex]\([0, 2\pi)\)[/tex], the solutions to the equation [tex]\(2 \cos^2 x + \sin x - 2 = 0\)[/tex] are:
[tex]\[
x = 0, \frac{\pi}{6}, \frac{5\pi}{6}
\][/tex]
Thus, the solutions are:
[tex]\[
\boxed{0, \frac{\pi}{6}, \frac{5\pi}{6}}
\][/tex]
