Question

Find all the solutions to

− 2cot^2x+4=16−6cot^2x
.

Assume that n
is any integer.





x=7π/6+2πn and x=11π/6+2πn
x=π/6+πn
x=π/6+2πn and x=7π/6+2πn
x=π/6+πn and x=7π/6+πn
There are no solutions.
x=π/6+πn and x=5π/6+πn

Answer 1

To solve the equation

\[
-2 \cot^2 x + 4 = 16 - 6 \cot^2 x,
\]

let's first rearrange the equation by moving all terms involving \(\cot^2 x\) to one side:

\[
-2 \cot^2 x + 6 \cot^2 x = 16 - 4.
\]

This simplifies to:

\[
4 \cot^2 x = 12.
\]

Next, we can divide both sides by 4:

\[
\cot^2 x = 3.
\]

Now, taking the square root of both sides gives us:

\[
\cot x = \pm \sqrt{3}.
\]

The general solutions for \(\cot x = \sqrt{3}\) occur at:

\[
x = \frac{\pi}{6} + \pi n \quad \text{(for } n \in \mathbb{Z}\text{)},
\]

and the general solutions for \(\cot x = -\sqrt{3}\) occur at:

\[
x = \frac{5\pi}{6} + \pi n \quad \text{(for } n \in \mathbb{Z}\text{)}.
\]

Therefore, combining both cases, we have:

1. \(x = \frac{\pi}{6} + \pi n\)
2. \(x = \frac{5\pi}{6} + \pi n\)

Thus, the complete set of solutions is:

\[
x = \frac{\pi}{6} + \pi n \quad \text{and} \quad x = \frac{5\pi}{6} + \pi n \quad \text{for } n \in \mathbb{Z}.
\]

The answer to your multiple-choice question is:

**\(x = \frac{\pi}{6} + \pi n\) and \(x = \frac{5\pi}{6} + \pi n\)**.