Math / Trigonometry

QuestionFind all solutions of the given equation. (Enter your answers as a comma-separated list. Let $k$ be any integer. Do not round coefficients of $k$ . Other terms can be rounded to two decimal places where appropriate.) $\begin{array}{r} 2 \cos (\theta)-\sqrt{3}=0 \\ \theta=\square \mathrm{rad} \end{array}$

Studdy Solution

STEP 1

1. The equation $2 \cos(\theta) - \sqrt{3} = 0$ is a trigonometric equation.
2. We are looking for all solutions in terms of $\theta$ , expressed in radians.
3. The general solution for cosine equations involves periodicity, specifically $2\pi$ .

STEP 2

1. Isolate $\cos(\theta)$ .
2. Solve for $\theta$ using the inverse cosine function.
3. Determine the general solution for $\theta$ considering the periodic nature of the cosine function.

STEP 3

First, isolate $\cos(\theta)$ by adding $\sqrt{3}$ to both sides and then dividing by 2:
$2 \cos(\theta) - \sqrt{3} = 0$ $2 \cos(\theta) = \sqrt{3}$ $\cos(\theta) = \frac{\sqrt{3}}{2}$

STEP 4

Find the principal values of $\theta$ for which $\cos(\theta) = \frac{\sqrt{3}}{2}$ . The cosine function equals $\frac{\sqrt{3}}{2}$ at:
$\theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{11\pi}{6}$

STEP 5

Since cosine is periodic with a period of $2\pi$ , the general solutions for $\theta$ are:
$\theta = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2k\pi$
where $k$ is any integer.
The solutions for $\theta$ are:
$\theta = \frac{\pi}{6} + 2k\pi, \frac{11\pi}{6} + 2k\pi$

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