# How do you solve trigonometric equations by the quadratic formula?

Mar 25, 2015

The quadratic formula can be useful in solving trigonometric (or other) kinds of equation. But strictly speaking, the quadratic formula is used only to solve quadratic equations.

Here is an example:

We can use the quadratic formula to solve

$2 {x}^{2} - 6 x + 3 = 0$.

We get $x = \frac{- \left(- 6\right) \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(2\right) \left(3\right)}}{2 \left(2\right)} = \frac{3 \pm \sqrt{3}}{2}$

Now what if we needed to solve: ${t}^{6} - 2 {t}^{3} - 2 = 0$?
This is not a quadratic equation.
But, if we substitute $x$ in place of ${t}^{3}$, we get ${x}^{2} - 2 x - 2 = 0$ Which we can solve by the quadratic formula, as above.
But we want $t$, not $x$. That's OK, we have gained this information:

${t}^{3} = \frac{3 \pm \sqrt{3}}{2}$

And (taking 3rd roots on both sides) gives us

$t = \sqrt{\frac{3 \pm \sqrt{3}}{2}}$ . . (No $\pm$ is needed for odd roots)

Now, suppose we need to solve ${\sin}^{2} t - 2 \sin t - 2 = 0$

This is a trigonometric equation, not a quadratic equation. Or is it?

Can't we "turn it into" a quadratic by substituting? (Such equations are sometimes called "quadratic in form".)

Let $x = \sin t$, we get our old friend ${x}^{2} - 2 x - 2 = 0$.

So $x = \sin t = \frac{3 \pm \sqrt{3}}{2}$

Now we need to solve $\sin t = \frac{3 + \sqrt{3}}{2}$ . find $t$.
That's going to be a problem because $\frac{3 + \sqrt{3}}{2}$ is greater than 1.

Solve $\sin t = \frac{3 - \sqrt{3}}{2}$ .

In this case there is a solution, but it is not one of the special angles. Using tables or electronics, we can get the reference angle ${39.3}^{\circ}$ Or the radian angle (the real number) $0.68668$.

In degrees the solutions are:
$t = {39.3}^{\circ} + {360}^{\circ} k$ for any integer $k$
$t = {140.7}^{\circ} + {360}^{\circ} k$ for any integer $k$