# Question 3efc5

Feb 10, 2017

θ = 90°, or 180°, or 270°.

#### Explanation:

First, we can factor our a 4cosθ to get:

4cosθ(cosθ + 1) = 0

This means that either 4cosθ = 0 => cosθ = 0

or cosθ + 1 = 0 => cosθ = -1

You can consult the trigonometric circle below to help:

graph{x^2 +y^2 = 1 [-2.5, 2.5, -1.25, 1.25]}

cosθ corresponds to the $x$ axis. The angle θ is measured from the Ox semiaxis, going counterclockwise. Therefore,

cosθ = 0 => θ = 90° or θ = 270°
cosθ = -1 => θ = 180°

So the three possible solutions are

{90°, 180°, 270°}#

Feb 10, 2017

$\theta = {90}^{o} , {180}^{o} \mathmr{and} {270}^{o}$

#### Explanation:

Factor: $4 \cos \theta \left(\cos \theta + 1\right) = 0$

$4 \cos \theta = 0 , \mathmr{and} \cos \theta = - 1$

From a trig. circle: $\cos \theta = 0$ occurs at ${90}^{o} \mathmr{and} {270}^{o}$
From a trig. circle: $\cos \theta = - 1$ occurs at ${180}^{o}$