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

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Answer 1 · 2015-05-15T14:25:01

On the trig unit circle:

$cos (2 x) = - \frac{3}{6} = - \frac{1}{2} \to 2 x = \frac{2 π}{3} and 2 x = \frac{4 π}{3}$ $cos (2x) = -3/6 = -1/2 -> 2x = (2pi)/3 and 2x = (4pi)/3$

Answer:
Within period $(0, 2 π) : x = \frac{π}{6} and x = 4 \frac{π}{6} = \frac{2 π}{3}$ $(0, 2pi): x = pi/6 and x = 4pi/6 = (2pi)/3$

Check:

$x = \frac{π}{6} \to 2 x = \frac{π}{3}; cos 2 x = \frac{1}{2} \to f (x) = - 3 + 3 = 0$ $x = pi/6 -> 2x = pi/3 ; cos 2x = 1/2 -> f(x) = -3 + 3 = 0$ OK

$x = 2 \frac{π}{3} \to 2 x = 4 \frac{π}{3} \to f (x) = 6 (- \frac{1}{2}) + 3 = 0$ $x = 2pi/3 -> 2x = 4pi/3 -> f(x) = 6(-1/2) + 3 = 0$ . OK

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