Find All Complex Solutions -cos(theta)^2+sin(theta)^2=cos(theta)+1

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

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

Subtract $\cos (θ)$ from both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

Simplify $- \cos^{2} (θ) + \sin^{2} (θ) - \cos (θ) - 1$ .

Move $- 1$ .

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

Reorder $\sin^{2} (θ)$ and $- 1$ .

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

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

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

Factor $- 1$ out of $\sin^{2} (θ)$ .

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

Factor $- 1$ out of $- 1 (1) - 1 (- \sin^{2} (θ))$ .

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

Rewrite $- 1 (1 - \sin^{2} (θ))$ as $- (1 - \sin^{2} (θ))$ .

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

Apply pythagorean identity.

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

Subtract $\cos^{2} (θ)$ from $- \cos^{2} (θ)$ .

$- 2 \cos^{2} (θ) - \cos (θ) = 0$

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

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

$- \cos (θ) (2 \cos (θ)) - \cos (θ) = 0$

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

$- \cos (θ) (2 \cos (θ)) - \cos (θ) \cdot 1 = 0$

Factor $- \cos (θ)$ out of $- \cos (θ) (2 \cos (θ)) - \cos (θ) \cdot 1$ .

$- \cos (θ) (2 \cos (θ) + 1) = 0$

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

$\cos (θ) = 0$

$2 \cos (θ) + 1 = 0$

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

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Solve $\cos (θ) = 0$ for $θ$ .

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

$θ = \arccos (0)$

Simplify the right side.

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

$θ = \frac{π}{2}$

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.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

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.

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

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

Set $2 \cos (θ) + 1$ equal to $0$ .

$2 \cos (θ) + 1 = 0$

Solve $2 \cos (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$2 \cos (θ) = - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$\cos (θ) = - \frac{1}{2}$

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

$θ = \arccos (- \frac{1}{2})$

Simplify the right side.

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

$θ = \frac{2 π}{3}$

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.

$θ = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

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.

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

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

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

Find All Complex Solutions -cos(theta)^2+sin(theta)^2=cos(theta)+1

Related Questions

Support LCMGCF.com by disabling your adblocker.