How do you solve the equation #2cos^2x- 7cosx + 3 = 0#?

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Answer 1 · 2016-08-30T17:53:22

#x = pi/3 + 2pin and x = (5pi)/3 + 2pin#

Solve by factoring.

#2cos^2x - 6cosx - cosx + 3 =0#

#2cosx(cosx - 3) - 1(cosx - 3) = 0#

#(2cosx -1)(cosx - 3) = 0#

#cosx= 1/2 and cosx = 3#

#x = pi/3 + 2pin, x = (5pi)/3 + 2pin#

Hopefully this helps!