Sure, let's solve the equation:
[tex]\[
\cot^2(x) + 3 \csc(x) = -3
\][/tex]
Step 1: Use trigonometric identities. Recall that [tex]\(\cot(x) = \frac{\cos(x)}{\sin(x)}\)[/tex] and [tex]\(\csc(x) = \frac{1}{\sin(x)}\)[/tex]. Therefore, we can rewrite the equation in terms of sine and cosine:
[tex]\[
\left(\frac{\cos(x)}{\sin(x)}\right)^2 + 3 \frac{1}{\sin(x)} + 3 = 0
\][/tex]
Step 2: Simplify the equation. Let [tex]\( u = \csc(x) = \frac{1}{\sin(x)} \)[/tex]. Then, [tex]\(\cot^2(x) = \csc^2(x) - 1\)[/tex], so the equation becomes:
[tex]\[
(u^2 - 1) + 3u + 3 = 0
\][/tex]
Step 3: Combine like terms:
[tex]\[
u^2 + 3u + 2 = 0
\][/tex]
Step 4: Factor the quadratic equation:
[tex]\[
(u + 1)(u + 2) = 0
\][/tex]
Step 5: Solve for [tex]\( u \)[/tex]:
[tex]\[
u + 1 = 0 \quad \text{or} \quad u + 2 = 0
\][/tex]
[tex]\[
u = -1 \quad \text{or} \quad u = -2
\][/tex]
Since [tex]\( u = \csc(x) = \frac{1}{\sin(x)} \)[/tex], we solve for [tex]\(\sin(x)\)[/tex]:
1. For [tex]\( u = -1 \)[/tex]:
[tex]\[
\frac{1}{\sin(x)} = -1 \implies \sin(x) = -1
\][/tex]
2. For [tex]\( u = -2 \)[/tex]:
[tex]\[
\frac{1}{\sin(x)} = -2 \implies \sin(x) = -\frac{1}{2}
\][/tex]
Step 6: Determine the angles in the interval [tex]\([0, 2\pi)\)[/tex] for these sine values.
1. For [tex]\(\sin(x) = -1\)[/tex]:
[tex]\[
x = \frac{3\pi}{2}
\][/tex]
2. For [tex]\(\sin(x) = -\frac{1}{2}\)[/tex]:
[tex]\[
x = \frac{7\pi}{6}, \frac{11\pi}{6}
\][/tex]
Therefore, the solutions to the equation [tex]\(\cot^2(x) + 3\csc(x) = -3\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[
x = \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{3\pi}{2}
\][/tex]
