Solve for θ cos(theta)=1/3

$\cos (θ) = \frac{1}{3}$

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

$θ = \arccos (\frac{1}{3})$

Simplify the right side.

Evaluate $\arccos (\frac{1}{3})$ .

$θ = 1.23095941$

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.

$θ = 2 (3.14159265) - 1.23095941$

Simplify $2 (3.14159265) - 1.23095941$ .

Multiply $2$ by $3.14159265$ .

$θ = 6.2831853 - 1.23095941$

Subtract $1.23095941$ from $6.2831853$ .

$θ = 5.05222588$

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.

$θ = 1.23095941 + 2 π n, 5.05222588 + 2 π n$ , for any integer $n$

Solve for θ cos(theta)=1/3

Related Questions

Support LCMGCF.com by disabling your adblocker.