How do you find all exact solutions on #(0,2pi)# of #2 cos^2(t)+3 cos(t)=-1# ?

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#2 cos^2(t)+3 cos(t)=-1#

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Answer 1 · 2018-06-13T19:55:29

#t = (2 pi)/3, pi, (4 pi)/3#

Given: #2 cos^2 t + 3cos t = -1#

#2 cos^2 t + 3cos t+1 = 0#

For some it is hard to factor trigonometric functions. Doing a substitution can help the process.

Let #a = cos t: " "2a^2 + 3a + 1 = 0#

Factor: #" "(2a + 1)(a + 1) = 0#

Reverse the substitution: #" "(2cos t + 1) (cos t + 1) = 0#

#2 cos t = -1; " " cos t = -1#

#cos t = -1/2#

From a unit circle the only place that #x = -1# is at #pi#.

Cosine is negative in the 2nd and 3rd quadrants. The reference angle is #pi/3# for #cos t = -1/2#

2nd quadrant: #t = pi - pi/3 = (3 pi)/3 - pi/3 = (2 pi)/3#

3rd quadrant: #t = pi + pi/3 = (3 pi)/3 + pi/3 = (4 pi)/3#