To solve the equation [tex]\( 2 \cos^2 x + 3 \cos x + 1 = 0 \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex], let's follow a detailed step-by-step process:
1. Substitute [tex]\(\cos x\)[/tex] with a new variable:
Let [tex]\( u = \cos x \)[/tex]. This substitution reduces the original trigonometric equation to an algebraic quadratic equation:
[tex]\[
2u^2 + 3u + 1 = 0
\][/tex]
2. Solve the quadratic equation:
The quadratic equation [tex]\( 2u^2 + 3u + 1 = 0 \)[/tex] can be solved using the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 1 \)[/tex].
Calculate the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1
\][/tex]
Find the roots:
[tex]\[
u = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} = \frac{-3 \pm 1}{4}
\][/tex]
This results in two solutions:
[tex]\[
u_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}
\][/tex]
[tex]\[
u_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1
\][/tex]
Therefore, [tex]\(\cos x = -\frac{1}{2}\)[/tex] or [tex]\(\cos x = -1\)[/tex].
3. Solve for [tex]\( x \)[/tex] based on [tex]\(\cos x\)[/tex]:
- For [tex]\(\cos x = -\frac{1}{2}\)[/tex]:
[tex]\[
x = \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
These are the principal values in the interval [tex]\([0, 2\pi)\)[/tex].
- For [tex]\(\cos x = -1\)[/tex]:
[tex]\[
x = \pi
\][/tex]
This is the principal value in the interval [tex]\([0, 2\pi)\)[/tex].
4. Combine the solutions:
Collecting all values, we get:
[tex]\[
x = \pi, \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
5. Check provided options:
The correct option is:
[tex]\[
D. \quad x = \pi, \frac{2\pi}{3}, \frac{4\pi}{3}
\][/tex]
Thus, the solutions to the equation [tex]\(2 \cos^2 x + 3 \cos x + 1 = 0\)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are [tex]\( x = \pi, \frac{2\pi}{3}, \frac{4\pi}{3}\)[/tex], corresponding to option D.
