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

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Answer 1 · 2015-05-15T14:25:01

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

How do you solve 6cos(2x) + 3 = 06cos(2x) + 3 = 0?