# Question #52d16

Sep 14, 2017

$x = {58}^{\circ} , {122}^{\circ}$

#### Explanation:

We have: $2 {\sin}^{2} \left(x\right) + 3 \sin \left(x\right) = 4$; $\left[0 , {360}^{\circ}\right)$

If you take a look at the equation, you may notice that it is almost in the form of a quadratic equation.

Let's subtract $4$ from both sides of the equation:

$R i g h t a r r o w 2 {\sin}^{2} \left(x\right) + 3 \sin \left(x\right) - 4 = 0$

We can consider $\sin \left(x\right)$ to be a variable of its own, so let's apply the quadratic formula:

$R i g h t a r r o w \sin \left(x\right) = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(2\right) \left(- 4\right)}}{2 \left(2\right)}$

$R i g h t a r r o w \sin \left(x\right) = \frac{- 3 \pm \sqrt{9 + 32}}{4}$

$R i g h t a r r o w \sin \left(x\right) = \frac{- 3 \pm \sqrt{41}}{4}$

$R i g h t a r r o w \sin \left(x\right) = \frac{- 3 \pm 6.40}{4}$

$R i g h t a r r o w \sin \left(x\right) = - 2.35 , 0.85$

However, the range of $\sin \left(x\right)$ is $\left[- 1 , 1\right]$.

So $\sin \left(x\right) = 0.85$.

Let's set the reference angle as $\sin \left(x\right) = 0.85 R i g h t a r r o w x = {58}^{\circ}$:

$R i g h t a r r o w \sin \left(x\right) = 0.85$

$R i g h t a r r o w x = {58}^{\circ} , {180}^{\circ} - {58}^{\circ}$

$\therefore x = {58}^{\circ} , {122}^{\circ}$

Therefore, the solutions to the equation are $x = {58}^{\circ}$ and $x = {122}^{\circ}$.