Math / Trigonometry

Question $\cot x+\sqrt{3}=0$

Studdy Solution

STEP 1

What is this asking? We need to find all the angles $x$ that make the equation $\cot x + \sqrt{3} = 0$ true. Watch out! Remember that the cotangent function repeats itself, so there will be infinitely many solutions!
Also, don't forget about those pesky domain restrictions!

STEP 2

1. Rewrite in terms of $\tan x$ .
2. Find the principal solution.
3. Find the general solution.

STEP 3

Alright, let's isolate $\cot x$ by subtracting $\sqrt{3}$ from both sides of the equation: $\cot x + \sqrt{3} - \sqrt{3} = 0 - \sqrt{3}$ $\cot x = -\sqrt{3}$

STEP 4

Now, since $\cot x$ is the reciprocal of $\tan x$ , we can rewrite this as: $\frac{1}{\tan x} = -\sqrt{3}$ Taking the reciprocal of both sides gives us: $\tan x = -\frac{1}{\sqrt{3}}$ Cool!

STEP 5

Let's make that fraction look a little nicer by rationalizing the denominator.
We'll multiply both the numerator and denominator by $\sqrt{3}$ : $\tan x = -\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$ Much better!

STEP 6

We know that $\tan(\frac{\pi}{6}) = \frac{\sqrt{3}}{3}$ , so our reference angle is $\frac{\pi}{6}$ .
Since the tangent is negative, our angle $x$ must be in either the second or fourth quadrant.
The principal solution (the one between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ ) is $x = -\frac{\pi}{6}$ .
Awesome!

STEP 7

Since the tangent function has a period of $\pi$ , we can find all the solutions by adding integer multiples of $\pi$ to our principal solution.

STEP 8

So, the general solution is given by: $x = -\frac{\pi}{6} + n\pi$ where $n$ is any integer.
Fantastic!

STEP 9

The solutions to the equation $\cot x + \sqrt{3} = 0$ are given by $x = -\frac{\pi}{6} + n\pi$ , where $n$ is any integer.

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