Find All Complex Solutions 4sin(x)^2=5+4cos(x)

$4 \sin^{2} (x) = 5 + 4 \cos (x)$

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

Subtract $5$ from both sides of the equation.

$4 \sin^{2} (x) - 5 = 4 \cos (x)$

Subtract $4 \cos (x)$ from both sides of the equation.

$4 \sin^{2} (x) - 5 - 4 \cos (x) = 0$

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

$4 (1 - \cos^{2} (x)) - 5 - 4 \cos (x) = 0$

Simplify each term.

Apply the distributive property.

$4 \cdot 1 + 4 (- \cos^{2} (x)) - 5 - 4 \cos (x) = 0$

Multiply $4$ by $1$ .

$4 + 4 (- \cos^{2} (x)) - 5 - 4 \cos (x) = 0$

Multiply $- 1$ by $4$ .

$4 - 4 \cos^{2} (x) - 5 - 4 \cos (x) = 0$

Subtract $5$ from $4$ .

$- 4 \cos^{2} (x) - 1 - 4 \cos (x) = 0$

Reorder the polynomial.

$- 4 \cos^{2} (x) - 4 \cos (x) - 1 = 0$

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

$- 4 {(u)}^{2} - 4 u - 1 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 4 u^{2} - 4 u - 1$ .

Factor $- 1$ out of $- 4 u^{2}$ .

$- (4 u^{2}) - 4 u - 1 = 0$

Factor $- 1$ out of $- 4 u$ .

$- (4 u^{2}) - (4 u) - 1 = 0$

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

$- (4 u^{2}) - (4 u) - 1 \cdot 1 = 0$

Factor $- 1$ out of $- (4 u^{2}) - (4 u)$ .

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

Factor $- 1$ out of $- (4 u^{2} + 4 u) - 1 (1)$ .

$- (4 u^{2} + 4 u + 1) = 0$

Factor using the perfect square rule.

Rewrite $4 u^{2}$ as ${(2 u)}^{2}$ .

$- ({(2 u)}^{2} + 4 u + 1) = 0$

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

$- ({(2 u)}^{2} + 4 u + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot (2 u) \cdot 1$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 u$ and $b = 1$ .

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

Divide each term in $- {(2 u + 1)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(2 u + 1)}^{2} = 0$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide ${(2 u + 1)}^{2}$ by $1$ .

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

Simplify the right side.

Divide $0$ by $- 1$ .

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

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

$2 u + 1 = 0$

Solve 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.

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

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

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

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 $n$

Find All Complex Solutions 4sin(x)^2=5+4cos(x)

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