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Solve each equation for all values of [tex]$x$[/tex].

[tex]\[ 3 \cot^2 x - 5 = -4 \][/tex]

A. [tex]\[ \frac{\pi}{3} + n\pi, \frac{2\pi}{3} + n\pi \][/tex]

B. [tex]\[ \frac{\pi}{6} + n\pi, \frac{5\pi}{6} + n\pi \][/tex]

C. [tex]\[ \frac{\pi}{3} + 2n\pi, \frac{2\pi}{3} + 2n\pi \][/tex]

D. [tex]\[ \frac{\pi}{3} + \frac{n\pi}{2}, \frac{2\pi}{3} + \frac{n\pi}{2} \][/tex]



Answer :

GinnyAnswer
To solve the trigonometric equation [tex]\(3 \cot^2 x - 5 = -4\)[/tex] for all values of [tex]\(x\)[/tex], follow these steps:

1. Isolate the trigonometric function:

Start with the given equation:
[tex]\[ 3 \cot^2 x - 5 = -4 \][/tex]

Add 5 to both sides to isolate the term involving [tex]\(\cot^2 x\)[/tex]:
[tex]\[ 3 \cot^2 x = 1 \][/tex]

2. Solve for [tex]\(\cot^2 x\)[/tex]:

Divide both sides by 3:
[tex]\[ \cot^2 x = \frac{1}{3} \][/tex]

3. Take the square root of both sides:

The cotangent function can be positive or negative, so we consider both square roots:
[tex]\[ \cot x = \pm \frac{1}{\sqrt{3}} \][/tex]

Simplify the expression:
[tex]\[ \cot x = \pm \frac{\sqrt{3}}{3} \][/tex]

4. Find the general solutions for [tex]\(x\)[/tex]:

The cotangent function equals [tex]\(\frac{\sqrt{3}}{3}\)[/tex] at specific angles. We know that:
[tex]\[ \cot x = \frac{\sqrt{3}}{3} \quad \text{at} \quad x = \frac{\pi}{6} + n\pi \quad \text{for integer} \ n \][/tex]
[tex]\[ \cot x = -\frac{\sqrt{3}}{3} \quad \text{at} \quad x = \frac{5\pi}{6} + n\pi \quad \text{for integer} \ n \][/tex]

5. Combine the general solutions:

Putting the two together, we get the general solutions for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\pi}{6} + n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + n\pi \][/tex]

Therefore, the solutions to the trigonometric equation [tex]\(3 \cot^2 x - 5 = -4\)[/tex] are:

[tex]\[ x = \frac{\pi}{6} + n\pi \quad \text{and} \quad x = \frac{5\pi}{6} + n\pi \quad \text{for integer} \ n \][/tex]

These derive the [tex]\(x\)[/tex] values where the initial equation holds true.

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