How do you solve #cos^2(3theta)-5cos(3theta)+4=0#?
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"Question #19ea3"
Please see the explanation.
Please notice that the quadratic factors into:
#(cos(3theta) - 1)(cos(3theta) - 4) = 0#
Which implies
#cos(3theta) = 1 and cos(3theta) = 4#
However, #cos(3theta) = 4# must be discarded, because it exceeds the range of the cosine function.
This leaves:
#cos(3theta) = 1#
To make the cosine function disappear, take the inverse cosine of both sides:
#3theta = cos^-1(1)#
The inverse cosine of 1 is zero:
#3theta = 0#
But it repeats every integer multiple of #2pi# from #-oo# to #oo#:
#3theta = 0 + 2npi; n = ...,-3, -2, -1, 0, 1, 2, 3,...#
Divide by 3:
#theta = 2/3npi; n = ...,-3, -2, -1, 0, 1, 2, 3,...#
Call 3t = T, and solve the quadratic equation for cos T:
#cos^2 T - 5cos T + 4 = 0#
#D = d^2 = 25 - 16 = 9 #--> #d = +- 3#
There are 2 real roots:
#cos T = -b/(2a) +- d/(2a) = 5/2 +- 3/2#
cos T1 = 4 (rejected as > 1)
#cos T2 = (5 - 3)/2 = 1#
cos T = cos 3t = 1
Trig unit circle -->
a/ #3t = 0 + 2kpi # -->
#t = (2kpi)/3#
b/ #3t = 2pi + 2kpi#-->
#t = (2pi)/3 + (2kpi)/3#
General answer: #t = (2kpi)/3#
