# If sin^4theta+cos^4theta=1, find theta?

Sep 8, 2017

${\sin}^{4} \theta + {\cos}^{4} \theta = 1$

$\implies {\left({\sin}^{2} \theta + {\cos}^{2} \theta\right)}^{2} - 2 {\sin}^{2} \theta {\cos}^{2} \theta = 1$

$\implies {1}^{2} - 2 {\sin}^{2} \theta {\cos}^{2} \theta = 1$

$\implies 2 {\sin}^{2} \theta {\cos}^{2} \theta = 0$

$\implies \sin \theta \cos \theta = 0$

So
when $\sin \theta = 0$

$\implies \theta = n \pi \text{ where } n \in \mathbb{Z}$

when $\cos \theta = 0$

$\implies \theta = \frac{\left(2 n + 1\right) \pi}{2} \text{ where } n \in \mathbb{Z}$

combining we get

$\theta = \frac{k \pi}{2} \text{ where } k \in \mathbb{Z}$

Sep 8, 2017

$x = \frac{n \pi}{2}$, where $n$ is an integer.

#### Explanation:

let we consider,
${\sin}^{2} x + {\cos}^{2} x = 1$

${\left({\sin}^{2} x + {\cos}^{2} x\right)}^{2} = {1}^{2}$

${\sin}^{4} x + {\cos}^{4} x + 2 {\sin}^{2} x {\cos}^{2} x = 1$

${\sin}^{4} x + {\cos}^{4} x = 1 - 2 {\sin}^{2} x {\cos}^{2} x$

since, ${\sin}^{4} x + {\cos}^{4} x = 1$, then

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

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

$\frac{1}{2} \left(4 {\sin}^{2} x {\cos}^{2} x\right) = 0$

$\frac{1}{2} {\left(2 \sin x \cos x\right)}^{2} = 0$

$\to \left(2 \sin x \cos x\right) = \sin 2 x$, therefore

$\frac{1}{2} {\left(\sin 2 x\right)}^{2} = 0$ $\to$${\sin}^{2} 2 x = 0$

$\sin 2 x = 0$

$2 x = 0 , \pi , 2 \pi , 3 \pi , 4 \pi , \ldots$

$x = 0 , \frac{\pi}{2} , \pi , \frac{3 \pi}{2} , 2 \pi . \ldots$

$x = \frac{n \pi}{2}$, where $n$ is an integer.