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

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

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

Add $\sin (x)$ to both sides of the equation.

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

Subtract $1$ from both sides of the equation.

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

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

Move $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

Apply pythagorean identity.

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

Subtract $\sin^{2} (x)$ from $- \sin^{2} (x)$ .

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

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

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

$- \sin (x) (2 \sin (x)) + \sin (x) = 0$

Rewrite $\sin (x)$ as $1 \sin (x)$ .

$- \sin (x) (2 \sin (x)) + 1 \sin (x) = 0$

Factor $- \sin (x)$ out of $1 \sin (x)$ .

$- \sin (x) (2 \sin (x)) - \sin (x) \cdot - 1 = 0$

Factor $- \sin (x)$ out of $- \sin (x) (2 \sin (x)) - \sin (x) \cdot - 1$ .

$- \sin (x) (2 \sin (x) - 1) = 0$

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

$\sin (x) = 0$

$2 \sin (x) - 1 = 0$

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 $n$

Set $2 \sin (x) - 1$ equal to $0$ and solve for $x$ .

Set $2 \sin (x) - 1$ equal to $0$ .

$2 \sin (x) - 1 = 0$

Solve $2 \sin (x) - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 \sin (x) = 1$

Divide each term in $2 \sin (x) = 1$ by $2$ and simplify.

Divide each term in $2 \sin (x) = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

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 $n$

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

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

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

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