Find All Complex Solutions 2sin(x)^2+cos(x)-1=0

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

Replace the $2 \sin^{2} (x)$ with $2 (1 - \cos^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

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

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $1$ from $2$ .

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

Substitute $u$ for $\cos (x)$ .

$- 2 {(u)}^{2} + u + 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 2 u^{2} + u + 1$ .

Factor $- 1$ out of $- 2 u^{2}$ .

$- (2 u^{2}) + u + 1 = 0$

Factor $- 1$ out of $u$ .

$- (2 u^{2}) - 1 (- u) + 1 = 0$

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

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

Factor $- 1$ out of $- (2 u^{2}) - 1 (- u)$ .

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

Factor $- 1$ out of $- (2 u^{2} - u) - 1 (- 1)$ .

$- (2 u^{2} - u - 1) = 0$

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 $- u$ .

$- (2 u^{2} - u - 1) = 0$

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

$- (2 u^{2} + (1 - 2) u - 1) = 0$

Apply the distributive property.

$- (2 u^{2} + 1 u - 2 u - 1) = 0$

Multiply $u$ by $1$ .

$- (2 u^{2} + u - 2 u - 1) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- (2 u^{2} + u - 2 u - 1) = 0$

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

$- (u (2 u + 1) - (2 u + 1)) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

$- ((2 u + 1) (u - 1)) = 0$

Remove unnecessary parentheses.

$- (2 u + 1) (u - 1) = 0$

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

$2 u + 1 = 0$

$u - 1 = 0$

Set $2 u + 1$ equal to $0$ and solve for $u$ .

Set $2 u + 1$ equal to $0$ .

$2 u + 1 = 0$

Solve $2 u + 1 = 0$ for $u$ .

Subtract $1$ from both sides of the equation.

$2 u = - 1$

Divide each term in $2 u = - 1$ by $2$ and simplify.

Divide each term in $2 u = - 1$ by $2$ .

$\frac{2 u}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 u}{2} = \frac{- 1}{2}$

Divide $u$ by $1$ .

$u = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$u = - \frac{1}{2}$

Set $u - 1$ equal to $0$ and solve for $u$ .

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

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

$u = - \frac{1}{2}, 1$

Substitute $\cos (x)$ for $u$ .

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

Set up each of the solutions to solve for $x$ .

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

$\cos (x) = 1$

Solve for $x$ in $\cos (x) = - \frac{1}{2}$ .

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 $n$

Solve for $x$ in $\cos (x) = 1$ .

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 $n$

List all of the solutions.

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

Consolidate the answers.

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

Find All Complex Solutions 2sin(x)^2+cos(x)-1=0

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