Solve for θ 4cos(theta)^2=3

$4 \cos^{2} (θ) = 3$

Step 1

Divide each term in $4 \cos^{2} (θ) = 3$ by $4$ and simplify.

Divide each term in $4 \cos^{2} (θ) = 3$ by $4$ .

$\frac{4 \cos^{2} (θ)}{4} = \frac{3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 \cos^{2} (θ)}{4} = \frac{3}{4}$

Divide $\cos^{2} (θ)$ by $1$ .

$\cos^{2} (θ) = \frac{3}{4}$

Step 2

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$\cos (θ) = \pm \sqrt{\frac{3}{4}}$

Step 3

Simplify $\pm \sqrt{\frac{3}{4}}$ .

Rewrite $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$\cos (θ) = \pm \frac{\sqrt{3}}{\sqrt{4}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\cos (θ) = \pm \frac{\sqrt{3}}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$\cos (θ) = \pm \frac{\sqrt{3}}{2}$

Step 4

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$\cos (θ) = \frac{\sqrt{3}}{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$\cos (θ) = - \frac{\sqrt{3}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$\cos (θ) = \frac{\sqrt{3}}{2}, - \frac{\sqrt{3}}{2}$

Step 5

Set up each of the solutions to solve for $θ$ .

$\cos (θ) = \frac{\sqrt{3}}{2}$

$\cos (θ) = - \frac{\sqrt{3}}{2}$

Step 6

Solve for $θ$ in $\cos (θ) = \frac{\sqrt{3}}{2}$ .

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (\frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arccos (\frac{\sqrt{3}}{2})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{6}$

Simplify $2 π - \frac{π}{6}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$θ = 2 π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $2 π$ and $\frac{6}{6}$ .

$θ = \frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

Multiply $6$ by $2$ .

$θ = \frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$θ = \frac{11 π}{6}$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 7

Solve for $θ$ in $\cos (θ) = - \frac{\sqrt{3}}{2}$ .

Take the inverse cosine of both sides of the equation to extract $θ$ from inside the cosine.

$θ = \arccos (- \frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{\sqrt{3}}{2})$ is $\frac{5 π}{6}$ .

$θ = \frac{5 π}{6}$

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - \frac{5 π}{6}$

Simplify $2 π - \frac{5 π}{6}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$θ = 2 π \cdot \frac{6}{6} - \frac{5 π}{6}$

Combine fractions.

Combine $2 π$ and $\frac{6}{6}$ .

$θ = \frac{2 π \cdot 6}{6} - \frac{5 π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 6 - 5 π}{6}$

Simplify the numerator.

Multiply $6$ by $2$ .

$θ = \frac{12 π - 5 π}{6}$

Subtract $5 π$ from $12 π$ .

$θ = \frac{7 π}{6}$

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\cos (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n$ , for any integer $n$

Step 8

List all of the solutions.

$θ = \frac{π}{6} + 2 π n, \frac{11 π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{7 π}{6} + 2 π n$ , for any integer $n$

Step 9

Consolidate the solutions.

Consolidate $\frac{π}{6} + 2 π n$ and $\frac{7 π}{6} + 2 π n$ to $\frac{π}{6} + π n$ .

$θ = \frac{π}{6} + π n, \frac{11 π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Consolidate $\frac{11 π}{6} + 2 π n$ and $\frac{5 π}{6} + 2 π n$ to $\frac{5 π}{6} + π n$ .

$θ = \frac{π}{6} + π n, \frac{5 π}{6} + π n$ , for any integer $n$

Solve for θ 4cos(theta)^2=3

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