# How do you solve 2cosx+3=0?

Apr 15, 2018

$x = \left(2 k + 1\right) \pi \pm \ln \left(\frac{3 + \sqrt{5}}{2}\right) i \text{ }$ for any integer $k$

#### Explanation:

Given:

$2 \cos x + 3 = 0$

Subtracting $3$ from both sides and dividing by $2$ this becomes:

$\cos x = - \frac{3}{2}$

This is outside the range $\left[- 1 , 1\right]$ of $\cos x$ as a real valued function of real values. So there are no Real solutions.

Euler's formula tells us:

${e}^{i x} = \cos x + i \sin x$

Taking conjugate, we get :-

${e}^{- i x} = \cos x - i \sin x$

Hence on adding above two equations we get :-

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

So the given equation becomes:

${e}^{i x} + {e}^{- i x} + 3 = 0$

Multiplying by $4 {e}^{i x}$ and rearranging slightly:

$0 = 4 {\left({e}^{i x}\right)}^{2} + 12 \left({e}^{i x}\right) + 4$

$\textcolor{w h i t e}{0} = {\left(2 {e}^{i x}\right)}^{2} + 2 \left(2 {e}^{i x}\right) \left(3\right) + 9 - 5$

$\textcolor{w h i t e}{0} = {\left(2 {e}^{i x} + 3\right)}^{2} - {\left(\sqrt{5}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(2 {e}^{i x} + 3 - \sqrt{5}\right) \left(2 {e}^{i x} + 3 + \sqrt{5}\right)$

So:

${e}^{i x} = \frac{- 3 \pm \sqrt{5}}{2}$

So:

$i x = \ln \left(\frac{- 3 \pm \sqrt{5}}{2}\right) + 2 k \pi i$

$\textcolor{w h i t e}{i x} = \pm \ln \left(\frac{3 + \sqrt{5}}{2}\right) + \left(2 k + 1\right) \pi i$

for any integer $k$

So:

$x = \left(2 k + 1\right) \pi \pm \ln \left(\frac{3 + \sqrt{5}}{2}\right) i$

Apr 15, 2018

No solutions for $x \in \mathbb{R}$

#### Explanation:

$2 \cos x + 3 = 0$

Subtract $3$ from both sides and divide by 2:

$\cos x = - \frac{3}{2}$

This show that the equation has no real solutions. We know this because:

$- 1 \le \cos x \le 1$

The graph of $y = 2 \cos \left(x\right) + 3$ confirms this: