Question Solve for $\theta$ given the equation $5+4\cos\theta=6$ .

Studdy Solution

STEP 1

Assumptions
1. We are given the equation $5 + 4 \cos \theta = 6$ .
2. We need to solve for $\theta$ .

STEP 2

To isolate $\cos \theta$ , we need to move the constant term on the left side of the equation to the right side.
$4 \cos \theta = 6 - 5$

STEP 3

Now, perform the subtraction on the right side of the equation.
$4 \cos \theta = 1$

STEP 4

To solve for $\cos \theta$ , divide both sides of the equation by 4.
$\cos \theta = \frac{1}{4}$

STEP 5

Now, we need to find the value(s) of $\theta$ that satisfy the equation. Since $\cos \theta = \frac{1}{4}$ , we will use the inverse cosine function, also known as arccosine.
$\theta = \arccos\left(\frac{1}{4}\right)$

STEP 6

The arccosine function will give us the principal value of $\theta$ . However, since the cosine function is periodic, there are infinitely many solutions to the equation. The general solution for $\theta$ in degrees is given by:
$\theta = \arccos\left(\frac{1}{4}\right) + 360^\circ k \quad \text{and} \quad \theta = -\arccos\left(\frac{1}{4}\right) + 360^\circ k$
where $k$ is any integer.

STEP 7

If we are looking for solutions in radians, the general solution for $\theta$ is given by:
$\theta = \arccos\left(\frac{1}{4}\right) + 2\pi k \quad \text{and} \quad \theta = -\arccos\left(\frac{1}{4}\right) + 2\pi k$
where $k$ is any integer.

STEP 8

The problem does not specify whether the solution should be in degrees or radians, nor does it restrict the domain for $\theta$ . Therefore, the solutions provided in STEP_6 and STEP_7 are the general solutions for $\theta$ .
The value of $\theta$ that satisfies $5 + 4 \cos \theta = 6$ is:
$\theta = \arccos\left(\frac{1}{4}\right) + 360^\circ k \quad \text{or} \quad \theta = -\arccos\left(\frac{1}{4}\right) + 360^\circ k$
in degrees, or
$\theta = \arccos\left(\frac{1}{4}\right) + 2\pi k \quad \text{or} \quad \theta = -\arccos\left(\frac{1}{4}\right) + 2\pi k$
in radians, for any integer $k$ .

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