# How do you solve the following equation 6 sin^2 x - sin x -2 = 0 in the interval [0, 2pi]?

Jul 9, 2016

Solve the quadratic and use a little trig to find solutions of $x = \frac{7 \pi}{6}$, $x = \frac{11 \pi}{6}$, $x = 0.73$, and $x = 2.41$.

#### Explanation:

I don't know about you, but I definitely don't like to solve equations involving ${\sin}^{2} x$. It looks really messy. To make it a little neater, let's use the substitution $u = \sin x$. Our equation becomes:
$6 {u}^{2} - u - 2 = 0$

Ah...that's better. Now we have a quadratic equation that we know how to solve. Let's try to factor it, using the AC method. First, we multiply the ${\textcolor{w h i t e}{x}}^{2}$ term by the constant term - in this case, the $6$ in our equation by the $- 2$:
$6 \cdot - 2 = - 12$

Then, we find two numbers that multiply to $- 12$ and add to the middle term, in this case the $- 1$. We'll need to list the factors of $- 12$:

$\textcolor{w h i t e}{X I} \underline{\text{Factors of -12}}$
$- 1 \cdot 12$ or $1 \cdot - 12$
$- 2 \cdot 6$ or $2 \cdot - 6$
$- 3 \cdot 4$ or $3 \cdot - 4$

Which of these adds to $- 1$? Take a good look and you'll see $3 \cdot - 4 = - 12$ and $3 - 4 = - 1$. These numbers satisfy our conditions. The next step is breaking $- u$ into $3 u - 4 u$, which are the numbers that add to $- 1$:
$6 {u}^{2} + 3 u - 4 u - 2 = 0$

We can factor out a $3 u$ from the first pair and a $- 2$ from the second pair:
$3 u \textcolor{red}{\left(2 u + 1\right)} - 2 \textcolor{red}{\left(2 u + 1\right)} = 0$

Note the common factor $2 u + 1$; we can pull that out too:
$\textcolor{red}{\left(2 u + 1\right)} \left(3 u - 2\right) = 0$

Using the zero product property, we can set our two factors equal to $0$ and solve:
$2 u + 1 = 0$ and $3 u - 2 = 0$
$\to u = - \frac{1}{2}$ and $u = \frac{2}{3}$

Remember that $u = \sin x$, so:
$\sin x = - \frac{1}{2}$ and $\sin x = \frac{2}{3}$

Now we just have to solve this on $\left[0 , 2 \pi\right]$. So, when does $\sin x$ equal $- \frac{1}{2}$? The unit circle tells us that $\sin x = - \frac{1}{2}$ when $x = \frac{7 \pi}{6}$ and $x = \frac{11 \pi}{6}$, so those are two solutions.

How about $\sin x = \frac{2}{3}$? We'll need to use a calculator for that. Take the inverse sine of both sides:
${\sin}^{- 1} \left(\sin x\right) = {\sin}^{- 1} \left(\frac{2}{3}\right)$
$\to x = {\sin}^{- 1} \left(\frac{2}{3}\right)$

Using a calculator, we obtain the principal solution $x \approx 0.73$. However, there is another solution that the calculator won't give us - the solution in the second quadrant. To find that, we subtract $0.73$ from $\pi$:
$x = \pi - 0.73 \approx 2.41$

Thus our solutions are $x = \frac{7 \pi}{6}$, $x = \frac{11 \pi}{6}$, $x = 0.73$, and $x = 2.41$.