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

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Answer 1 · 2016-10-03T07:48:29

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

$x² - 3x + 2 = 0$

factors into:

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

x = 2 and x = 1

Substitute for x:

$2sin(t/10) = 2$ and $2sin(t/10) = 1$

Divide both by 2:

$sin(t/10) = 1$ and $sin(t/10) = 1/2$

Take the inverse sine of both, remembering that $sin(x) = 1/2$ is true in both the first and second quadrants:

$t/10 = pi/2$ , $t/10 = pi/6$ , and $t/10 = 5pi/6$

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

$t/10 = pi/2 + 2npi$ , $t/10 = pi/6 + 2npi$ , and $t/10 = 5pi/6 + 2npi$

Multiply by 10:

$t = 5pi + 20npi$ , $t = 5pi/3 + 20npi$ , and $t = 25pi/3 + 20npi$

The periodic function x = 2 sin ( t/10 )x=2sin(t10)x = 2 sin ( t/10 ) satisfies x^2-3x+2=0x2−3x+2=0x^2-3x+2=0. How do you find all the values of t?