# Find all solutions of the equation. 4\cos(\theta) + 1 = 0

## Question:

Find all solutions of the equation.

{eq}\displaystyle\; 4\cos(\theta) + 1 = 0 {/eq}

## Trigonometric Equation:

There are numerous occasions when we are presented with trigonometric expressions in which equality is established, but we do not know the value of the argument that satisfies such equality. In those cases, we must solve that trigonometric equation and obtain the value of that argument. Generally, we make some previous steps in order to adapt the equation and later to make use of the inverse trigonometric function that corresponds.

## Answer and Explanation:

{eq}\eqalign{ & {\text{Let's solve the trigonometric equation }}\,4\cos \theta + 1 = 0. \cr & {\text{Subtracting 1 on both sides of the equation}}: \cr & \,\,\,\,\,\,\,4\cos \theta = - 1 \cr & {\text{Dividing by 4 on both sides}}: \cr & \,\,\,\,\,\,\,\cos \theta = - \frac{1}{4} \cr & {\text{Using the inverse trigonometric function }}{\cos ^{ - 1}}\left( {\cos u} \right) = u{\text{ }} \cr & {\text{applying to both sides of the equation:}} \cr & \,\,\,\,\,\,\,{\cos ^{ - 1}}\left[ {\cos \left( \theta \right)} \right] = {\cos ^{ - 1}}\left[ { - \frac{1}{4}} \right] \cr & {\text{We have that }}\,{\text{if }}\,\cos v = u\,\, \Rightarrow v = {\cos ^{ - 1}}\left( u \right),{\text{ then solving for }}\,\theta {\text{:}} \cr & \,\,\,\,\,\,\theta = {\cos ^{ - 1}}\left( {\cos \left( \theta \right)} \right) = {\cos ^{ - 1}}\left( { - \frac{1}{4}} \right) \cr & {\text{So}}{\text{, we have that }}\,\cos \left( {1.823} \right) = - \frac{1}{4}{\text{ and }}\,\cos \left( { - 1.823} \right) = - \frac{1}{4}{\text{. Then:}} \cr & \,\,\,\,\,\,\,\theta = {\cos ^{ - 1}}\left( { - \frac{1}{4}} \right) = 1.823\,\,{\text{ and }}\,\,\theta = {\cos ^{ - 1}}\left( { - \frac{1}{4}} \right) = - 1.823 \cr & {\text{As the cosine function is periodic with a 2}}\pi {\text{ period}}{\text{, the solution }} \cr & {\text{is repeated every }}2\pi n{\text{:}} \cr & \,\,\,\,\,\,\,\theta = 1.823{\text{ + }}2\pi n{\text{ and }}\,\theta = - 1.823{\text{ + }}2\pi n \cr & {\text{Therefore}}{\text{, the solutions are: }}\,\boxed{\theta = 1.823{\text{ + }}2\pi n{\text{ and }}\,\theta = - 1.823{\text{ + }}2\pi n} \cr} {/eq}

#### Learn more about this topic:

from High School Precalculus Textbook

Chapter 22 / Lesson 5