# Find all solutions of the equation. \tan(\frac{\theta}{4}) + \sqrt{3} = 0

## Question:

Find all solutions of the equation.

{eq}\displaystyle\; \tan\left(\frac{\theta}{4}\right) + \sqrt{3} = 0 {/eq}

## Trigonometric Equations:

A trigonometric equations is an equation that has at least one trigonometric function. The three basic trigonometric functions are {eq}\text{sine(sin), tangent(tan), cosine(cos)} {/eq}. An example of a trigonometric equation is {eq}\sin^2x + \cos^2 x = 1 {/eq}.If a trigonometric equation holds for all the angles, then it ie refered to as a trigonometric equation which is an identity. The example we gave earlier holds for all the angles. Thus, it is a trigonometric identity.

## Answer and Explanation:

We have the equation:

- {eq}\tan\left(\dfrac{\theta}{4}\right) + \sqrt{3} = 0 {/eq}

Subtracting {eq}\sqrt{3} {/eq} from both sides of the equation, we get:

- {eq}\tan\left(\dfrac{\theta}{4}\right) = -\sqrt{3} {/eq}

Taking the inverse of {eq}\tan {/eq} on both sides of the equation, we get:

- {eq}\dfrac{\theta}{4} = \tan^{-1}(-\sqrt{3}) {/eq}

- {eq}\dfrac{\theta}{4} = -\dfrac{\pi}{3} + \pi = \dfrac{2\pi }{3} {/eq}

Since {eq}\tan {/eq} is negative in the second and in the fourth quadrant, then:

- {eq}\dfrac{\theta}{4} =\left( \dfrac{2\pi }{3}, \quad \dfrac{5\pi }{3}\right) {/eq}

Tangent has a period of {eq}\pi {/eq}. Therefore:

- {eq}\dfrac{\theta}{4} =\left( \dfrac{2\pi }{3} + \pi k, \quad \dfrac{5\pi }{3} + \pi k\right) {/eq}

Where {eq}k {/eq} is any integer.

Solving for {eq}\theta {/eq}, we get:

- {eq}\theta =\boxed{\color{blue}{4\left( \dfrac{2\pi }{3} + \pi k, \quad \dfrac{5\pi }{3} + \pi k\right)}} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from High School Precalculus Textbook

Chapter 22 / Lesson 5