# The periodic function x = 2 sin ( t/10 ) satisfies x^2-3x+2=0. How do you find all the values of t?

Oct 3, 2016

$t = 5 \pi + 20 n \pi$, $t = 5 \frac{\pi}{3} + 20 n \pi$, and $t = 25 \frac{\pi}{3} + 20 n \pi$

#### Explanation:

x² - 3x + 2 = 0

factors into:

(x - 2)(x - 1) = 0

x = 2 and x = 1

Substitute for x:

$2 \sin \left(\frac{t}{10}\right) = 2$ and $2 \sin \left(\frac{t}{10}\right) = 1$

Divide both by 2:

$\sin \left(\frac{t}{10}\right) = 1$ and $\sin \left(\frac{t}{10}\right) = \frac{1}{2}$

Take the inverse sine of both, remembering that $\sin \left(x\right) = \frac{1}{2}$ is true in both the first and second quadrants:

$\frac{t}{10} = \frac{\pi}{2}$, $\frac{t}{10} = \frac{\pi}{6}$, and $\frac{t}{10} = 5 \frac{\pi}{6}$

Account for t to cause the sine function to repeat n rotations:

$\frac{t}{10} = \frac{\pi}{2} + 2 n \pi$, $\frac{t}{10} = \frac{\pi}{6} + 2 n \pi$, and $\frac{t}{10} = 5 \frac{\pi}{6} + 2 n \pi$

Multiply by 10:

$t = 5 \pi + 20 n \pi$, $t = 5 \frac{\pi}{3} + 20 n \pi$, and $t = 25 \frac{\pi}{3} + 20 n \pi$