How do you solve #2cos^2x-3cosx+1=0# and find all solutions in the interval #(0,2pi)#?

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Answer 1 · 2016-10-24T14:37:55

Let #t = cosx#

We can thus rewrite the equation as #2t^2 - 3t + 1 = 0# .

This can be factored as follows:

#2t^2 - 2t - t + 1= 0#

#2t(t - 1) - 1(t - 1) = 0#

#(2t - 1)(t - 1) = 0#

#t = 1/2 and 1#

We can replace #t# with #cosx#.

#cosx = 1/2 and cosx = 1#

#x = pi/3, (5pi)/3#

Hopefully this helps!