# Question 265f1

Feb 14, 2018

$\frac{\pi}{3}$ or $\frac{5 \pi}{3}$

#### Explanation:

Use the double angle formula
$\cos \left(2 A\right) = 1 - 2 {\sin}^{2} A$
to rewrite the equation in the form
$0 = \sin \left(\frac{x}{2}\right) - \cos \left(x\right) = \sin \left(\frac{x}{2}\right) - 1 + 2 {\sin}^{2} \left(\frac{x}{2}\right)$

Substituting $t = \sin \left(\frac{x}{2}\right)$ this equation becomes
$2 {t}^{2} + t - 1 = 0 \implies \left(2 t - 1\right) \left(t + 1\right) = 0$

Since $x \in \left[0 , 2 \pi\right)$, we have $\frac{x}{2} \in \left[0 , \pi\right)$, so that $t = \sin \left(\frac{x}{2}\right) \ge 0$ Thus
$2 t - 1 = 0 \implies t = \frac{1}{2} \implies \sin \left(\frac{x}{2}\right) = \frac{1}{2}$
Thus, either $\frac{x}{2} = \frac{\pi}{6}$ or $\frac{x}{2} = \frac{5 \pi}{6}$

Feb 14, 2018

$x = \pm \pi$ or $x = \pm \frac{\pi}{2}$

#### Explanation:

Starting with $\sin \left(\frac{x}{2}\right) - \cos x = 0$, use the half angle formula for $\sin \left(\frac{x}{2}\right)$:
$\sin \left(\frac{x}{2}\right) = \sqrt{\frac{1 - \cos x}{2}}$ and rewrite the original equation:

$\sqrt{\frac{1 - \cos x}{2}} - \cos x = 0$
Add $\cos x$ to both sides: sqrt((1-cosx)/2)-cosx=0 =>

$\sqrt{\frac{1 - \cos x}{2}} = \cos x$
Square both sides to obtain $\frac{1 - \cos x}{2} = {\cos}^{2} x$
Multiply both sides by 2, then put everything on one side:
$1 - \cos x = 2 {\cos}^{2} x$
$2 {\cos}^{2} x + \cos x - 1 = 0$
You've essentially got a simple quadratic equation to factor now. If $n = \cos x$, you could rewrite it as $2 {\cos}^{2} x - \cos x + 1 = 0$
=> $2 {n}^{2} - n - 1$, which factors as $\left(2 n - 1\right) \left(n + 1\right)$.

So $2 {\cos}^{2} x + \cos x - 1 = 0$ => $\left(2 \cos x - 1\right) \left(\cos x + 1\right) = 0$

Like solving for the zeros in any quadratic equation, set each factor equal to zero to find the solutions:
$2 \cos x - 1 = 0$
$2 \cos x = 1$

We know from the unit circle that $\cos x = \frac{1}{2}$ at $x = \pm \frac{\pi}{3}$, or equivalently, $x = \pm {60}^{o}$

The next factor:
$\cos x + 1 = 0$
$\cos x = - 1$

Likewise, the unit circle shows us that $\cos x = - 1$ where $x = \pm \pi$, and that $\pm \pi = \pm {180}^{o}$

So, $x = \pm \pi$ or $x = \pm \frac{\pi}{3}$
Or in degrees: $x = \pm {180}^{o}$ or $x = {60}^{o}$