# How do you find the general solution for 2sinx + 2/sinx = 5?

Oct 3, 2015

$S = \left\{\frac{\pi}{6} + 2 \pi k , \frac{5 \pi}{6} + 2 \pi k\right\}$

or, in degrees,

S = {30º + 360kº, 150 + 360kº}

#### Explanation:

Multiply both sides by $\sin \left(x\right)$

$2 {\sin}^{2} \left(x\right) + 2 = 5 \sin \left(x\right)$

Replace $\sin \left(x\right)$ for $y$

$2 {y}^{2} + 2 = 5 y$

Isolate $y$

$2 {y}^{2} - 5 y + 2 = 0$

$y = \frac{5 \pm \sqrt{25 - 4 \cdot 2 \cdot 2}}{4} = \frac{5 \pm \sqrt{25 - 16}}{4}$
$y = \frac{5 \pm 3}{4}$

So we have that $y = 2$ or $y = \frac{1}{2}$

Since we're dealing with trig functions, we know that $- 1 < y < 1$, so the only valid answer is $y = \frac{1}{2}$ or

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

We know from the special angles / unit circle that there are 2 angles in which this true.

30º or $\frac{\pi}{6}$ and 150º or $\frac{5 \pi}{6}$

However, any of these angles plus or minus a full loop would give the same answer, so we say that set of solutions $S$ is

$S = \left\{\frac{\pi}{6} + 2 \pi k , \frac{5 \pi}{6} + 2 \pi k\right\}$

or, in degrees,

S = {30º + 360kº, 150 + 360kº}

Where $k$ is an integer constant. It's also important to remember that since we were dividing by a sine at first, we could make a not of the restriction that we can't have any value of $x$ such that $\sin \left(x\right) = 0$, so $x \ne \pi k$ or, in degrees x != 180kº where $k$ is an integer constant