Question

I need help solving this practice I’m having trouble with it

Answer 1

From the problem, we have :

$2\sin ^2x+5\sin x-3=0$

Let m = sin x

$2m^2+5m-3=0$

Factor completely :

$\begin{gathered} 2m^2+5m-3=0 \\ (2m-1)(m+3)=0 \end{gathered}$

Equate factors to 0 then solve for m :

$\begin{gathered} 2m-1=0 \\ 2m=1 \\ m=\frac{1}{2} \\ \\ m+3=0 \\ m=-3 \end{gathered}$

Bring back m = sin x :

$\begin{gathered} m=\frac{1}{2}\Rightarrow\sin x=\frac{1}{2} \\ m=-3\Rightarrow\sin x=-3 \end{gathered}$

Note that the sin of an angle is ranging from -1 to 1, so sin x = -3 is invalid. Therefore, neglect sin x = -3 and use sin x = 1/2 only.

$\begin{gathered} \sin x=\frac{1}{2} \\ \text{take arcsin :} \\ \arcsin (\sin x)=\frac{1}{2} \\ x=30\quad \text{and}\quad 150 \end{gathered}$

Convert 30 and 150 degrees to radians :

$\begin{gathered} 30\times\frac{\pi}{180}=\frac{\pi}{6} \\ 150\times\frac{\pi}{180}=\frac{5\pi}{6} \end{gathered}$

Note that 2π is one full circle.

So we can add it to the solution.

That will be :

$\begin{gathered} \frac{\pi}{6}+2\pi n \\ \text{and} \\ \frac{5\pi}{6}+2\pi n \end{gathered}$

ANSWERS :

$\begin{gathered} \frac{\pi}{6}+2\pi n,\quad \text{ where n is an integer} \\ \text{and} \\ \frac{5\pi}{6}+2\pi n,\quad \text{ where n is an integer} \end{gathered}$