The periodic function #x = 2 sin ( t/10 )# satisfies #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 = 5pi + 20npi#, #t = 5pi/3 + 20npi#, and #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#