Question

find all angles, $0^{\circ} \leq \theta < 360^{\circ}$, that solve the following equation.
$\cos \theta = \frac{\sqrt{2}}{2}$
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$\theta = \square^{\circ}$
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Explanation:

Step1: Recall cosine values

We know that $\cos\theta = \frac{\sqrt{2}}{2}$ is a known cosine value for special angles. The reference angle $\alpha$ where $\cos\alpha=\frac{\sqrt{2}}{2}$ is $45^{\circ}$, since $\cos45^{\circ}=\frac{\sqrt{2}}{2}$.

Step2: Determine quadrants

Cosine is positive in the first and fourth quadrants (because cosine corresponds to the x - coordinate on the unit circle, and x - coordinates are positive in the first and fourth quadrants).

Step3: Find angles in each quadrant

• First quadrant: The angle $\theta$ in the first quadrant with reference angle $45^{\circ}$ is $\theta = 45^{\circ}$.
• Fourth quadrant: To find the angle in the fourth quadrant, we use the formula $\theta=360^{\circ}-\alpha$, where $\alpha = 45^{\circ}$. So $\theta = 360^{\circ}- 45^{\circ}=315^{\circ}$.

Answer:

$45^{\circ}, 315^{\circ}$