# Question #96293

Jan 19, 2018

On the domain, $\left\{x \in R | 0 \le x \le 2 \pi\right\}$

$x = \frac{\pi}{6} , x = \frac{5 \pi}{6} , x = \frac{7 \pi}{6} , x = \frac{11 \pi}{6}$

Or, in degrees;

On the domain, $\left\{x \in R | {0}^{\circ} \le x \le {360}^{\circ}\right\}$

$x = {30}^{\circ} , x = {150}^{\circ} , x = {210}^{\circ} , x = {330}^{\circ}$

#### Explanation:

You are going to need to use the pythagorean identity (${\sin}^{2} x + {\cos}^{2} x = 1$) to solve this problem.

$7 {\sin}^{2} x + 3 {\cos}^{2} x = 4$

As, ${\sin}^{2} x + {\cos}^{2} x = 1$ then; ${\cos}^{2} x = 1 - {\sin}^{2} x$ which we can use to substitute into the given equation.

$7 {\sin}^{2} x + 3 \left(\textcolor{red}{1 - {\sin}^{2} x}\right) = 4$

$7 {\sin}^{2} x + 3 - 3 {\sin}^{2} x = 4$

Combining like terms;

$4 {\sin}^{2} x + 3 = 4$

Moving all of the information to the left of the equality yields;

$4 {\sin}^{2} x - 1 = 0$

The next clever piece of maths comes because you will need to factorise this expression. So what we do is substitute a variable in for the trigonometric ratio.

i.e. let $u = \sin x$

$\therefore 4 {\sin}^{2} x - 1 = 4 {u}^{2} - 1 = 0$

Which we can factorise using the difference of perfect squares.

$4 {u}^{2} - 1 = \left(2 u + 1\right) \left(2 u - 1\right) = 0$

By the null factor law we can conclude that $u = \pm \frac{1}{2}$

$\therefore \sin x = \pm \frac{1}{2}$

So, on the domain, $\left\{x \in R | 0 \le x \le 2 \pi\right\}$

$x = \frac{\pi}{6} , x = \frac{5 \pi}{6} , x = \frac{7 \pi}{6} , x = \frac{11 \pi}{6}$

Or, in degrees;

On the domain, $\left\{x \in R | {0}^{\circ} \le x \le {360}^{\circ}\right\}$

$x = {30}^{\circ} , x = {150}^{\circ} , x = {210}^{\circ} , x = {330}^{\circ}$

I hope that helps :)