To solve the equation [tex]\( 4 \cos^2 x = 5 + 4 \sin x \)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex], we can follow these steps:
1. Express [tex]\(\cos^2 x\)[/tex] in terms of [tex]\(\sin x\)[/tex]:
Using the Pythagorean identity [tex]\(\cos^2 x = 1 - \sin^2 x\)[/tex], substitute this into the given equation:
[tex]\[
4 (1 - \sin^2 x) = 5 + 4 \sin x
\][/tex]
2. Simplify the equation:
Distribute the 4 on the left-hand side and move everything to one side of the equation:
[tex]\[
4 - 4 \sin^2 x = 5 + 4 \sin x
\][/tex]
[tex]\[
-4 \sin^2 x + 4 - 5 - 4 \sin x = 0
\][/tex]
[tex]\[
-4 \sin^2 x - 4 \sin x - 1 = 0
\][/tex]
3. Rearrange the equation:
To make it easier to solve, we can rewrite it in the standard quadratic form:
[tex]\[
4 \sin^2 x + 4 \sin x + 1 = 0
\][/tex]
4. Solve the quadratic equation:
Let's solve for [tex]\(\sin x\)[/tex] using the quadratic formula [tex]\(ax^2 + bx + c = 0\)[/tex], which is [tex]\(\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]. Here, [tex]\(a = 4\)[/tex], [tex]\(b = 4\)[/tex], and [tex]\(c = 1\)[/tex]:
[tex]\[
\sin x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}
\][/tex]
[tex]\[
\sin x = \frac{-4 \pm \sqrt{16 - 16}}{8}
\][/tex]
[tex]\[
\sin x = \frac{-4 \pm \sqrt{0}}{8}
\][/tex]
[tex]\[
\sin x = \frac{-4 \pm 0}{8} = \frac{-4}{8} = -\frac{1}{2}
\][/tex]
5. Find the angles [tex]\(x\)[/tex] that satisfy [tex]\(\sin x = -\frac{1}{2}\)[/tex]:
The solutions in the interval [tex]\([0, 2\pi)\)[/tex] where [tex]\(\sin x = -\frac{1}{2}\)[/tex] are:
[tex]\[
x = \frac{7\pi}{6}, \quad x = \frac{11\pi}{6}
\][/tex]
Thus, the solutions to the equation [tex]\( 4 \cos^2 x = 5 + 4 \sin x \)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
\boxed{\frac{7\pi}{6}, \frac{11\pi}{6}}
\][/tex]
