Find All Complex Solutions 2+2sin(theta)=4cos(theta)^2

$2 + 2 \sin (θ) = 4 \cos^{2} (θ)$

Subtract $4 \cos^{2} (θ)$ from both sides of the equation.

$2 + 2 \sin (θ) - 4 \cos^{2} (θ) = 0$

Replace the $- 4 \cos^{2} (θ)$ with $- 4 (1 - \sin^{2} (θ))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$2 + 2 \sin (θ) - 4 (1 - \sin^{2} (θ)) = 0$

Simplify each term.

Apply the distributive property.

$2 + 2 \sin (θ) - 4 \cdot 1 - 4 (- \sin^{2} (θ)) = 0$

Multiply $- 4$ by $1$ .

$2 + 2 \sin (θ) - 4 - 4 (- \sin^{2} (θ)) = 0$

Multiply $- 1$ by $- 4$ .

$2 + 2 \sin (θ) - 4 + 4 \sin^{2} (θ) = 0$

Subtract $4$ from $2$ .

$2 \sin (θ) - 2 + 4 \sin^{2} (θ) = 0$

Reorder the polynomial.

$4 \sin^{2} (θ) + 2 \sin (θ) - 2 = 0$

Substitute $u$ for $\sin (θ)$ .

$4 {(u)}^{2} + 2 (u) - 2 = 0$

Factor the left side of the equation.

Factor $2$ out of $4 u^{2} + 2 u - 2$ .

Factor $2$ out of $4 u^{2}$ .

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

Factor $2$ out of $2 u$ .

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

Factor $2$ out of $- 2$ .

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

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

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

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

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

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 $b = 1$ .

Multiply by $1$ .

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

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

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 $u$ .

Add $1$ to 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 $1$ .

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

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

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

$u + 1 = 0$

Subtract $1$ from both sides of the equation.

$u = - 1$

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

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

Substitute $\sin (θ)$ for $u$ .

$\sin (θ) = \frac{1}{2}, - 1$

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

$\sin (θ) = \frac{1}{2}$

$\sin (θ) = - 1$

Solve for $θ$ in $\sin (θ) = \frac{1}{2}$ .

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

$θ = \arcsin (\frac{1}{2})$

Simplify the right side.

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$θ = π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

$θ = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

Find the period of $\sin (θ)$ .

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 $\sin (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Solve for $θ$ in $\sin (θ) = - 1$ .

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

$θ = \arcsin (- 1)$

Simplify the right side.

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

$θ = - \frac{π}{2}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{2} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$θ = 2 π + \frac{π}{2} + π - 2 π$

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$θ = \frac{3 π}{2}$

Find the period of $\sin (θ)$ .

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 $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

List the new angles.

$θ = \frac{3 π}{2}$

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

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

List all of the solutions.

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer $n$

Find All Complex Solutions 2+2sin(theta)=4cos(theta)^2

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