Solve for θ cos(3theta)=1

$\cos (3 θ) = 1$

Step 1

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

$3 θ = \arccos (1)$

Step 2

Simplify the right side.

The exact value of $\arccos (1)$ is $0$ .

$3 θ = 0$

Step 3

Divide each term in $3 θ = 0$ by $3$ and simplify.

Divide each term in $3 θ = 0$ by $3$ .

$\frac{3 θ}{3} = \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{0}{3}$

Divide $θ$ by $1$ .

$θ = \frac{0}{3}$

Simplify the right side.

Divide $0$ by $3$ .

$θ = 0$

Step 4

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.

$3 θ = 2 π - 0$

Step 5

Solve for $θ$ .

Simplify.

Multiply $- 1$ by $0$ .

$3 θ = 2 π + 0$

Add $2 π$ and $0$ .

$3 θ = 2 π$

Divide each term in $3 θ = 2 π$ by $3$ and simplify.

Divide each term in $3 θ = 2 π$ by $3$ .

$\frac{3 θ}{3} = \frac{2 π}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 θ}{3} = \frac{2 π}{3}$

Divide $θ$ by $1$ .

$θ = \frac{2 π}{3}$

Step 6

Find the period of $\cos (3 θ)$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Step 7

The period of the $\cos (3 θ)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

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

Step 8

Consolidate the answers.

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

Solve for θ cos(3theta)=1

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