Find All Complex Solutions 4sin(x)=-cos(x)^2+1

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

Step 1

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

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

$4 \sin (x) + \cos^{2} (x) = 1$

Subtract $1$ from both sides of the equation.

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

Step 2

Simplify $4 \sin (x) + \cos^{2} (x) - 1$ .

Reorder $\cos^{2} (x)$ and $- 1$ .

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

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

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

Factor $- 1$ out of $\cos^{2} (x)$ .

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

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (x))$ .

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

Rewrite $- 1 (1 - \cos^{2} (x))$ as $- (1 - \cos^{2} (x))$ .

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

Apply pythagorean identity.

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

Step 3

Factor $\sin (x)$ out of $4 \sin (x) - \sin^{2} (x)$ .

Factor $\sin (x)$ out of $4 \sin (x)$ .

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

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

$\sin (x) \cdot 4 + \sin (x) (- \sin (x)) = 0$

Factor $\sin (x)$ out of $\sin (x) \cdot 4 + \sin (x) (- \sin (x))$ .

$\sin (x) (4 - \sin (x)) = 0$

Step 4

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

$\sin (x) = 0$

$4 - \sin (x) = 0$

Step 5

Set $\sin (x)$ equal to $0$ and solve for $x$ .

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Solve $\sin (x) = 0$ for $x$ .

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

$x = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

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.

$x = π - 0$

Subtract $0$ from $π$ .

$x = π$

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.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

Set $4 - \sin (x)$ equal to $0$ and solve for $x$ .

Set $4 - \sin (x)$ equal to $0$ .

$4 - \sin (x) = 0$

Solve $4 - \sin (x) = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$- \sin (x) = - 4$

Divide each term in $- \sin (x) = - 4$ by $- 1$ and simplify.

Divide each term in $- \sin (x) = - 4$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{- 4}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{- 4}{- 1}$

Divide $\sin (x)$ by $1$ .

$\sin (x) = \frac{- 4}{- 1}$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$\sin (x) = 4$

The range of sine is $- 1 \leq y \leq 1$ . Since $4$ does not fall in this range, there is no solution.

No solution

Step 7

The final solution is all the values that make $\sin (x) (4 - \sin (x)) = 0$ true.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Find All Complex Solutions 4sin(x)=-cos(x)^2+1

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