Solve for θ 1+cos(theta)=3cos(theta)

$1 + \cos (θ) = 3 \cos (θ)$

Step 1

Move all terms containing $\cos (θ)$ to the left side of the equation.

Subtract $3 \cos (θ)$ from both sides of the equation.

$1 + \cos (θ) - 3 \cos (θ) = 0$

Subtract $3 \cos (θ)$ from $\cos (θ)$ .

$1 - 2 \cos (θ) = 0$

Step 2

Subtract $1$ from both sides of the equation.

$- 2 \cos (θ) = - 1$

Step 3

Divide each term in $- 2 \cos (θ) = - 1$ by $- 2$ and simplify.

Divide each term in $- 2 \cos (θ) = - 1$ by $- 2$ .

$\frac{- 2 \cos (θ)}{- 2} = \frac{- 1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 \cos (θ)}{- 2} = \frac{- 1}{- 2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$\cos (θ) = \frac{1}{2}$

Step 4

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

$θ = \arccos (\frac{1}{2})$

Step 5

Simplify the right side.

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 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.

$θ = 2 π - \frac{π}{3}$

Step 7

Simplify $2 π - \frac{π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Step 8

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 $1$ .

$2 π$

Step 9

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

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Solve for θ 1+cos(theta)=3cos(theta)

Related Questions

Support LCMGCF.com by disabling your adblocker.