Math / Trigonometry

Question[-/1 Points] DETAILS MY NOTES SPRECALC8 7.4.032.
Find all solutions of the given equation. (Enter your answers as a comma-separated list. Let $k$ be any integer. Do not round coefficients $\begin{array}{l} \sqrt{3} \cot (\theta)+1=0 \\ \theta=\square \end{array}$ $\square$ Submit Ahswer

Studdy Solution

STEP 1

What is this asking? Find all the angles, $\theta$ , that make this equation, $\sqrt{3} \cot (\theta) + 1 = 0$ , true! Watch out! Remember, $\cot(\theta)$ is the reciprocal of $\tan(\theta)$ , so $\cot(\theta) = \frac{1}{\tan(\theta)}$ !
Also, there are going to be infinitely many solutions, so we'll use $k$ to represent any integer.

STEP 2

1. Isolate the cotangent function
2. Relate cotangent to tangent
3. Find the principal solutions
4. Express the general solution

STEP 3

We start with our equation: $\sqrt{3} \cot (\theta) + 1 = 0$ We want to isolate $\cot(\theta)$ , so let's subtract $1$ from both sides: $\sqrt{3} \cot (\theta) = -1$

STEP 4

Now, we divide both sides by $\sqrt{3}$ to get $\cot(\theta)$ all by itself: $\cot (\theta) = \frac{-1}{\sqrt{3}}$ Woohoo! Cotangent is isolated!

STEP 5

Since $\cot(\theta) = \frac{1}{\tan(\theta)}$ , we can rewrite our equation as: $\frac{1}{\tan(\theta)} = \frac{-1}{\sqrt{3}}$ To solve for $\tan(\theta)$ , we can take the reciprocal of both sides: $\tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3}$ Fantastic! Now we have a tangent equation!

STEP 6

We're looking for an angle whose tangent is $-\sqrt{3}$ .
We know $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ .
Since our tangent is negative, our angle $\theta$ will be in the second and fourth quadrants. The reference angle is $\frac{\pi}{3}$ .

STEP 7

In the second quadrant, $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . In the fourth quadrant, $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ . These are our principal solutions!

STEP 8

Since the tangent function has a period of $\pi$ , we can find all solutions by adding integer multiples of $\pi$ to our principal solutions. So, our general solutions are: $\theta = \frac{2\pi}{3} + k\pi$ $\theta = \frac{5\pi}{3} + k\pi$ where $k$ is any integer.
Amazing!

STEP 9

$\theta = \frac{2\pi}{3} + k\pi, \frac{5\pi}{3} + k\pi$

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Question[-/1 Points] DETAILS MY NOTES SPRECALC8 7.4.032.
Find all solutions of the given equation. (Enter your answers as a comma-separated list. Let $k$ be any integer. Do not round coefficients $\begin{array}{l} \sqrt{3} \cot (\theta)+1=0 \\ \theta=\square \end{array}$ $\square$ Submit Ahswer

Studdy Solution

STEP 1

STEP 2

1. Isolate the cotangent function
2. Relate cotangent to tangent
3. Find the principal solutions
4. Express the general solution

STEP 3

We start with our equation: $\sqrt{3} \cot (\theta) + 1 = 0$ We want to isolate $\cot(\theta)$ , so let's subtract $1$ from both sides: $\sqrt{3} \cot (\theta) = -1$

STEP 4

Now, we divide both sides by $\sqrt{3}$ to get $\cot(\theta)$ all by itself: $\cot (\theta) = \frac{-1}{\sqrt{3}}$ Woohoo! Cotangent is isolated!

STEP 5

Since $\cot(\theta) = \frac{1}{\tan(\theta)}$ , we can rewrite our equation as: $\frac{1}{\tan(\theta)} = \frac{-1}{\sqrt{3}}$ To solve for $\tan(\theta)$ , we can take the reciprocal of both sides: $\tan(\theta) = \frac{\sqrt{3}}{-1} = -\sqrt{3}$ Fantastic! Now we have a tangent equation!

STEP 6

We're looking for an angle whose tangent is $-\sqrt{3}$ .
We know $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ .
Since our tangent is negative, our angle $\theta$ will be in the second and fourth quadrants. The reference angle is $\frac{\pi}{3}$ .

STEP 7

In the second quadrant, $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . In the fourth quadrant, $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ . These are our principal solutions!

STEP 8

Since the tangent function has a period of $\pi$ , we can find all solutions by adding integer multiples of $\pi$ to our principal solutions. So, our general solutions are: $\theta = \frac{2\pi}{3} + k\pi$ $\theta = \frac{5\pi}{3} + k\pi$ where $k$ is any integer.
Amazing!

STEP 9

$\theta = \frac{2\pi}{3} + k\pi, \frac{5\pi}{3} + k\pi$

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Studdy Solution

STEP 1

STEP 2

1. Isolate the cotangent function2. Relate cotangent to tangent3. Find the principal solutions4. Express the general solution

STEP 3

We **start** with our equation: 3cot⁡(θ)+1=0 \sqrt{3} \cot (\theta) + 1 = 0 3​cot(θ)+1=0 We want to **isolate** cot⁡(θ)\cot(\theta)cot(θ), so let's **subtract** 111 from both sides: 3cot⁡(θ)=−1 \sqrt{3} \cot (\theta) = -1 3​cot(θ)=−1

STEP 4

Now, we **divide** both sides by 3\sqrt{3}3​ to get cot⁡(θ)\cot(\theta)cot(θ) all by itself: cot⁡(θ)=−13 \cot (\theta) = \frac{-1}{\sqrt{3}} cot(θ)=3​−1​ Woohoo! Cotangent is isolated!

STEP 5

STEP 6

We're looking for an angle whose tangent is −3-\sqrt{3}−3​.We know tan⁡(π3)=3\tan\left(\frac{\pi}{3}\right) = \sqrt{3}tan(3π​)=3​.Since our tangent is negative, our angle θ\thetaθ will be in the second and fourth quadrants. The **reference angle** is π3\frac{\pi}{3}3π​.

STEP 7

In the second quadrant, θ=π−π3=2π3\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}θ=π−3π​=32π​. In the fourth quadrant, θ=2π−π3=5π3\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}θ=2π−3π​=35π​. These are our **principal solutions**!

STEP 8

STEP 9

θ=2π3+kπ,5π3+kπ\theta = \frac{2\pi}{3} + k\pi, \frac{5\pi}{3} + k\piθ=32π​+kπ,35π​+kπ

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1. Isolate the cotangent function
2. Relate cotangent to tangent
3. Find the principal solutions
4. Express the general solution

We start with our equation: $\sqrt{3} \cot (\theta) + 1 = 0$ We want to isolate $\cot(\theta)$ , so let's subtract $1$ from both sides: $\sqrt{3} \cot (\theta) = -1$

Now, we divide both sides by $\sqrt{3}$ to get $\cot(\theta)$ all by itself: $\cot (\theta) = \frac{-1}{\sqrt{3}}$ Woohoo! Cotangent is isolated!

We're looking for an angle whose tangent is $-\sqrt{3}$ .
We know $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ .
Since our tangent is negative, our angle $\theta$ will be in the second and fourth quadrants. The reference angle is $\frac{\pi}{3}$ .

In the second quadrant, $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ . In the fourth quadrant, $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$ . These are our principal solutions!

$\theta = \frac{2\pi}{3} + k\pi, \frac{5\pi}{3} + k\pi$