Find All Complex Solutions cos(x)^2-cos(x)=0

$\cos^{2} (x) - \cos (x) = 0$

Step 1

Factor $\cos (x)$ out of $\cos^{2} (x) - \cos (x)$ .

Factor $\cos (x)$ out of $\cos^{2} (x)$ .

$\cos (x) \cos (x) - \cos (x) = 0$

Factor $\cos (x)$ out of $- \cos (x)$ .

$\cos (x) \cos (x) + \cos (x) \cdot - 1 = 0$

Factor $\cos (x)$ out of $\cos (x) \cos (x) + \cos (x) \cdot - 1$ .

$\cos (x) (\cos (x) - 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cos (x) = 0$

$\cos (x) - 1 = 0$

Step 3

Set $\cos (x)$ equal to $0$ and solve for $x$ .

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Solve $\cos (x) = 0$ for $x$ .

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

$x = \arccos (0)$

Simplify the right side.

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

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.

$x = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

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 $1$ .

$2 π$

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

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

Step 4

Set $\cos (x) - 1$ equal to $0$ and solve for $x$ .

Set $\cos (x) - 1$ equal to $0$ .

$\cos (x) - 1 = 0$

Solve $\cos (x) - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$\cos (x) = 1$

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

$x = \arccos (1)$

Simplify the right side.

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

$x = 0$

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.

$x = 2 π - 0$

Subtract $0$ from $2 π$ .

$x = 2 π$

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 $1$ .

$2 π$

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 5

The final solution is all the values that make $\cos (x) (\cos (x) - 1) = 0$ true.

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

Find All Complex Solutions cos(x)^2-cos(x)=0

Related Questions

Support LCMGCF.com by disabling your adblocker.