# Question d55d1

Nov 16, 2017

$x = \left(2 n + 1\right) \left(\cdot \frac{\pi}{4}\right)$ where $n \to \mathbb{Z}$

#### Explanation:

${\cos}^{2} x - {\sin}^{2} x = 0$

or, $\cos 2 x = 0$

or, $2 x = \left(2 n + 1\right) \cdot \left(\frac{\pi}{2}\right)$

or, $x = \left(2 n + 1\right) \left(\cdot \frac{\pi}{4}\right)$ where $n \to \mathbb{Z}$

Nov 16, 2017

$x = n {360}^{\circ} \pm {45}^{\circ} \mathmr{and} 2 n \pi \pm \frac{\pi}{4} , n \in \mathbb{Z}$

#### Explanation:

The double angle formula gives us $\cos \left(2 A\right) = {\cos}^{2} A - {\sin}^{2} A$

Using this, we know that $\cos \left(2 x\right) = 0$

$2 x = \arccos \left(0\right) = {90}^{\circ} \mathmr{and} \frac{\pi}{2}$

$x = {45}^{\circ} \mathmr{and} \frac{\pi}{4}$

However, $x = n {360}^{\circ} \pm {45}^{\circ} \mathmr{and} 2 n \pi \pm \frac{\pi}{4} , n \in \mathbb{Z}$

Nov 16, 2017

x = 45 degree

#### Explanation:

We know ${\cos}^{2} x + {\sin}^{2} x = 1. \mathmr{and} , {\cos}^{2} x = 1 - {\sin}^{2} x$

Put the value of ${\cos}^{2} x$, in ${\cos}^{2} x - {\sin}^{2} x = 0$ we get,

$\Rightarrow 1 - {\sin}^{2} x - {\sin}^{2} x = 0$

$\Rightarrow - 2 {\sin}^{2} x = - 1$

$\Rightarrow {\sin}^{2} x = \frac{1}{2}$

$\Rightarrow \sin x = \sqrt{\frac{1}{2}} = \sin {45}^{\cdot}$

$\Rightarrow x = {45}^{\cdot}$