Find All Complex Solutions -2cos(x)^2=-cos(x)-1

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

Step 1

Move all the expressions to the left side of the equation.

Add $\cos (x)$ to both sides of the equation.

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

Add $1$ to both sides of the equation.

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

Step 2

Factor the left side of the equation.

Factor $- 1$ out of $- 2 \cos^{2} (x) + \cos (x) + 1$ .

Factor $- 1$ out of $- 2 \cos^{2} (x)$ .

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

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

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

Rewrite $1$ as $- 1 (- 1)$ .

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

Factor $- 1$ out of $- (2 \cos^{2} (x)) - 1 (- \cos (x))$ .

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

Factor $- 1$ out of $- (2 \cos^{2} (x) - \cos (x)) - 1 (- 1)$ .

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

Factor.

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

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

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

Rewrite $- 1$ as $1$ plus $- 2$

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

Apply the distributive property.

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

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

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

Factor the polynomial by factoring out the greatest common factor, $2 \cos (x) + 1$ .

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

Remove unnecessary parentheses.

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

Step 3

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

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

$\cos (x) - 1 = 0$

Step 4

Set $2 \cos (x) + 1$ equal to $0$ and solve for $x$ .

Set $2 \cos (x) + 1$ equal to $0$ .

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

Solve $2 \cos (x) + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 \cos (x) = - 1$

Divide each term in $2 \cos (x) = - 1$ by $2$ and simplify.

Divide each term in $2 \cos (x) = - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$x = \arccos (- \frac{1}{2})$

Simplify the right side.

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

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.

$x = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

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.

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 5

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 6

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

$x = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, 2 π n, 2 π + 2 π n$ , for any integer $n$

Find All Complex Solutions -2cos(x)^2=-cos(x)-1

Related Questions

Support LCMGCF.com by disabling your adblocker.