How do you solve $4 {cos}^{2} x - 1 = 0$ $4cos^2x-1=0$ ?

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Answer 1 · 2016-10-21T01:05:00

$x = \pm \frac{4 π}{3}, \pm \frac{2 π}{3}, \pm \frac{5 π}{3}, \pm \frac{π}{3}$ $x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3$

This is a trivial quadratic equation in $cos x$ $cosx$

$4 {cos}^{2} x - 1 = 0$ $4cos^2x-1=0$
$:. 4cos^2x = 1$
$:. cos^2x = 1/4$
$:. cosx = +-1/2$

In order to solve this you need to be familiar with the graphs of the trig functions along with common values:

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The range of soutions is not specified in the question, so lets assume we want a solutio in radians, and $-2pi <= x <= 2pi$

Consider, $cos=1/2 => x=-2pi+pi/3,-pi/3,pi/3, 2pi-pi/3$
$:. x=(-5pi)/3,(-pi)/3,pi/3, (5pi)/3$

Consider, $cos=-1/2 => x=-2pi+(2pi)/3,(-2pi)/3,(2pi)/3, 2pi-(2pi)/3$
$:. x=(-4pi)/3,(-2pi)/3,(2pi)/3, (4pi)/3$

So the solutions are:
$:. x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3$

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