# How do you solve 2cosx+1=0 algebraically?

Jun 14, 2018

$\frac{2 \pi}{3} , \frac{4 \pi}{3}$ on $0 \le x \le 2 \pi$

#### Explanation:

$2 \cos x + 1 = 0$

$2 \cos x = - 1$

$\cos x = - \frac{1}{2}$

${\cos}^{-} 1 \left(- \frac{1}{2}\right) = x$

Use the unit circle to solve:

${\cos}^{-} 1 \left(- \frac{1}{2}\right) = \frac{2 \pi}{3} , \frac{4 \pi}{3}$

Jun 14, 2018

Given:

$2 \cos \left(x\right) + 1 = 0$

Subtract 1 from both sides of the equation:

$2 \cos \left(x\right) + 1 \textcolor{red}{- 1} = 0 \textcolor{red}{- 1}$

$2 \cos \left(x\right) = - 1$

Divide both sides of the equation by 2:

$\frac{2}{\textcolor{red}{2}} \cos \left(x\right) = - \frac{1}{\textcolor{red}{2}}$

$\cos \left(x\right) = - \frac{1}{2}$

Use the inverse cosine function on both sides:

$\textcolor{red}{{\cos}^{-} 1} \left(\cos \left(x\right)\right) = \textcolor{red}{{\cos}^{-} 1} \left(- \frac{1}{2}\right)$

$x = {\cos}^{-} 1 \left(- \frac{1}{2}\right)$

Within the domain $0 \le x < 2 \pi$ the above equation is true for 2 values of x:

$x = \frac{2}{3} \pi \mathmr{and} x = \frac{4}{3} \pi$

To represent all of the possible solutions, we add integer multiples of $2 \pi$ to both answers:

$x = \frac{2}{3} \pi + 2 n \pi \mathmr{and} x = \frac{4}{3} \pi + 2 n \pi , n \in \mathbb{Z}$