Solve for x 1-cos(x)=0

$1 - \cos (x) = 0$

Step 1

Subtract $1$ from both sides of the equation.

$- \cos (x) = - 1$

Step 2

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

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

$\frac{- \cos (x)}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\cos (x)}{1} = \frac{- 1}{- 1}$

Divide $\cos (x)$ by $1$ .

$\cos (x) = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$\cos (x) = 1$

Step 3

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

$x = \arccos (1)$

Step 4

Simplify the right side.

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

$x = 0$

Step 5

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.

$x = 2 π - 0$

Step 6

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 7

Find the period of $\cos (x)$ .

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 8

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

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

Step 9

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Solve for x 1-cos(x)=0

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