# How do you solve cos^2 theta=2cos^2 (theta/2) in the domain 0<=theta<=360?

Jun 26, 2018

The solutuions are $S = \left\{{128.2}^{\circ} , {231.8}^{\circ}\right\}$

#### Explanation:

We need

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

$- 1 \le \cos \theta \le 1$

Then, the equation is

${\cos}^{2} \theta = 2 {\cos}^{2} \left(\frac{\theta}{2}\right)$

${\cos}^{2} \theta = \cos \theta + 1$

${\cos}^{2} \theta - \cos \theta - 1 = 0$

This is a quadratic equation in ${\cos}^{2} \theta$

The solutions are

$\cos \theta = \frac{1 \pm \sqrt{{\left(- 1\right)}^{2} - 4 \left(1\right) \left(- 1\right)}}{2}$

$= \frac{1 \pm \sqrt{5}}{2}$

$\left\{\begin{matrix}\cos \theta = 1.62 \\ \cos \theta = - 0.618\end{matrix}\right.$

$\implies$, {(theta=O/),(theta=128.2^@ ; 231.8^@):}

$\setminus \theta = {128.18}^{\setminus} \circ , {231.82}^{\setminus} \circ$

#### Explanation:

Given that

$\setminus {\cos}^{2} \setminus \theta = 2 \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2}$

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

$4 \setminus {\cos}^{4} \setminus \frac{\setminus \theta}{2} + 1 - 4 \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = 2 \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2}$

$4 \setminus {\cos}^{4} \setminus \frac{\setminus \theta}{2} - 6 \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} + 1 = 0$

Solving above quadratic equation in terms of $\setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2}$ using quadratic formula as follows

$\setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = \setminus \frac{- \left(- 6\right) \setminus \pm \setminus \sqrt{{\left(- 6\right)}^{2} - 4 \left(4\right) \left(1\right)}}{2 \left(4\right)}$

$\setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = \setminus \frac{3 \setminus \pm \setminus \sqrt{5}}{4}$
But , $0 \setminus \le \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} \setminus \le 1$

$\setminus \therefore \setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = \setminus \frac{3 - \setminus \sqrt{5}}{4}$

$\setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = {\left(\setminus \frac{\setminus \sqrt{5} - 1}{2 \setminus \sqrt{2}}\right)}^{2}$

Let $\setminus {\cos}^{2} \setminus \frac{\setminus \theta}{2} = \setminus {\cos}^{2} \setminus \alpha$

$\setminus \frac{\setminus \theta}{2} = n \setminus \pi \setminus \pm \setminus \alpha$

$\setminus \theta = 2 \left(n \setminus \pi \setminus \pm \setminus \alpha\right)$

Where, $\setminus \cos \setminus \alpha = \setminus \frac{\setminus \sqrt{5} - 1}{2 \setminus \sqrt{2}} = {64.09}^{\setminus} \circ$ &
$n$ is any integer i.e. $n = 0 , \setminus \pm 1 , \setminus \pm 2 , \setminus \pm 3 , \setminus \ldots$

But since, $0 \setminus \le \setminus \theta \setminus \le {360}^{\setminus} \circ$ hence, setting n=0\ &\ 1 & taking positive & negative signs respectively, we get two values of $\setminus \theta$ as follows

$\setminus \theta = 2 \left(0 + {64.09}^{\setminus} \circ\right) , 2 \left({180}^{\setminus} \circ - {64.09}^{\setminus} \circ\right)$
$= {128.18}^{\setminus} \circ , {231.82}^{\setminus} \circ$