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How do you solve #6cos(2x) + 3 = 0#?

1 Answer

May 15, 2015

On the trig unit circle:

#cos (2x) = -3/6 = -1/2 -> 2x = (2pi)/3 and 2x = (4pi)/3#

Answer:
Within period #(0, 2pi): x = pi/6 and x = 4pi/6 = (2pi)/3#

Check:

#x = pi/6 -> 2x = pi/3 ; cos 2x = 1/2 -> f(x) = -3 + 3 = 0# OK

#x = 2pi/3 -> 2x = 4pi/3 -> f(x) = 6(-1/2) + 3 = 0#. OK

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