Solve for θ cos(theta)-1=0

$\cos (θ) - 1 = 0$

Add $1$ to both sides of the equation.

$\cos (θ) = 1$

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

$θ = \arccos (1)$

Simplify the right side.

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

$θ = 0$

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.

$θ = 2 π - 0$

Subtract $0$ from $2 π$ .

$θ = 2 π$

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.

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

Consolidate the answers.

$θ = 2 π n$ , for any integer $n$

Solve for θ cos(theta)-1=0

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