Algebra Examples

$\cos^{3} (x) - 2 \cos^{2} (x) - \cos (x) + 2 = 0$

Step 1

Factor $\cos^{3} (x) - 2 \cos^{2} (x) - \cos (x) + 2$ .

Tap for more steps...

Step 1.1

Factor out the greatest common factor from each group.

Tap for more steps...

Step 1.1.1

Group the first two terms and the last two terms.

$\cos^{3} (x) - 2 \cos^{2} (x) - \cos (x) + 2 = 0$

Step 1.1.2

Factor out the greatest common factor (GCF) from each group.

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

Step 1.2

Factor the polynomial by factoring out the greatest common factor, $\cos (x) - 2$ .

$(\cos (x) - 2) (\cos^{2} (x) - 1) = 0$

Step 1.3

Rewrite $1$ as $1^{2}$ .

$(\cos (x) - 2) (\cos^{2} (x) - 1^{2}) = 0$

Step 1.4

Factor.

Tap for more steps...

Step 1.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (x)$ and $b = 1$ .

$(\cos (x) - 2) ((\cos (x) + 1) (\cos (x) - 1)) = 0$

Step 1.4.2

Remove unnecessary parentheses.

$(\cos (x) - 2) (\cos (x) + 1) (\cos (x) - 1) = 0$

Step 2

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

$\cos (x) - 2 = 0$

$\cos (x) + 1 = 0$

$\cos (x) - 1 = 0$

Step 3

Set $\cos (x) - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.1

Set $\cos (x) - 2$ equal to $0$ .

$\cos (x) - 2 = 0$

Step 3.2

Solve $\cos (x) - 2 = 0$ for $x$ .

Tap for more steps...

Step 3.2.1

Add $2$ to both sides of the equation.

$\cos (x) = 2$

Step 3.2.2

The range of cosine is $- 1 \leq y \leq 1$ . Since $2$ does not fall in this range, there is no solution.

No solution

Step 4

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

Tap for more steps...

Step 4.1

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

$\cos (x) + 1 = 0$

Step 4.2

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

Tap for more steps...

Step 4.2.1

Subtract $1$ from both sides of the equation.

$\cos (x) = - 1$

Step 4.2.2

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

$x = \arccos (- 1)$

Step 4.2.3

Simplify the right side.

Tap for more steps...

Step 4.2.3.1

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

$x = π$

Step 4.2.4

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 4.2.5

Subtract $π$ from $2 π$ .

$x = π$

Step 4.2.6

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

Tap for more steps...

Step 4.2.6.1

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

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.7

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

$x = π + 2 π n$ , for any integer $n$

Step 5

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

Tap for more steps...

Step 5.1

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

$\cos (x) - 1 = 0$

Step 5.2

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

Tap for more steps...

Step 5.2.1

Add $1$ to both sides of the equation.

$\cos (x) = 1$

Step 5.2.2

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

$x = \arccos (1)$

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

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

$x = 0$

Step 5.2.4

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.

$x = 2 π - 0$

Step 5.2.5

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 5.2.6

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

Tap for more steps...

Step 5.2.6.1

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

$\frac{2 π}{| b |}$

Step 5.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.7

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 6

The final solution is all the values that make $(\cos (x) - 2) (\cos (x) + 1) (\cos (x) - 1) = 0$ true.

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

Step 7

Consolidate the answers.

$x = π n$ , for any integer $n$