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

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

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

Add $2 \cos^{2} (x)$ to both sides of the equation.

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

Subtract $4$ from both sides of the equation.

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

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

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

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $4$ from $2$ .

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

Reorder the polynomial.

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

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

$- 2 {(u)}^{2} + 5 (u) - 2 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- 2 u^{2} + 5 u - 2$ .

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

$- (2 u^{2}) + 5 u - 2 = 0$

Factor $- 1$ out of $5 u$ .

$- (2 u^{2}) - (- 5 u) - 2 = 0$

Rewrite $- 2$ as $- 1 (2)$ .

$- (2 u^{2}) - (- 5 u) - 1 \cdot 2 = 0$

Factor $- 1$ out of $- (2 u^{2}) - (- 5 u)$ .

$- (2 u^{2} - 5 u) - 1 \cdot 2 = 0$

Factor $- 1$ out of $- (2 u^{2} - 5 u) - 1 (2)$ .

$- (2 u^{2} - 5 u + 2) = 0$

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 $a \cdot c = 2 \cdot 2 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 u$ .

$- (2 u^{2} - 5 u + 2) = 0$

Rewrite $- 5$ as $- 1$ plus $- 4$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Remove unnecessary parentheses.

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

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

$2 u - 1 = 0$

$u - 2 = 0$

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 $u - 2$ equal to $0$ and solve for $u$ .

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Add $2$ to both sides of the equation.

$u = 2$

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

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

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

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

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

$\sin (x) = \frac{1}{2}$

$\sin (x) = 2$

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

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

$x = \arcsin (\frac{1}{2})$

Simplify the right side.

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

$x = \frac{π}{6}$

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.

$x = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Find the period of $\sin (x)$ .

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

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

Solve for $x$ in $\sin (x) = 2$ .

The range of sine is $- 1 \leq y \leq 1$ . Since $2$ does not fall in this range, there is no solution.

No solution

List all of the solutions.

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

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

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