How do you solve $1 = 8 cos (2 x + 1) - 3$ $1=8cos(2x +1) -3$ ?

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Answer 1 · 2015-05-13T20:04:09

First add 3 to both sides to get:

$4 = 8 cos (2 x + 1)$ $4 = 8 cos(2x+1)$

Next, divide both sides by 8 to get:

$\frac{1}{2} = cos (2 x + 1)$ $1/2 = cos(2x+1)$

If $cos (2 x + 1) = \frac{1}{2}$ $cos(2x+1) = 1/2$ then $2 x + 1 = 2 n π \pm \frac{π}{3}$ $2x+1 = 2n pi +- pi/3$ where $n$ $n$ is any integer.

Subtract 1 from both sides to get:

$2 x = 2 n π \pm \frac{π}{3} - 1$ $2x = 2n pi +- pi/3 - 1$

Divide both sides by 2 to get:

$x = n π \pm \frac{π}{6} - \frac{1}{2}$ $x = n pi +- pi/6 - 1/2$

How do you solve 1=8cos(2x +1) -31=8cos(2x+1)−31=8cos(2x +1) -3?