Solve for θ sin(theta)^2-sin(theta)=0

$\sin^{2} (θ) - \sin (θ) = 0$

Factor the left side of the equation.

Let $u = \sin (θ)$ . Substitute $u$ for all occurrences of $\sin (θ)$ .

$u^{2} - u = 0$

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

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

$u \cdot u - u = 0$

Factor $u$ out of $- u$ .

$u \cdot u + u \cdot - 1 = 0$

Factor $u$ out of $u \cdot u + u \cdot - 1$ .

$u (u - 1) = 0$

Replace all occurrences of $u$ with $\sin (θ)$ .

$\sin (θ) (\sin (θ) - 1) = 0$

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

$\sin (θ) = 0$

$\sin (θ) - 1 = 0$

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

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Solve $\sin (θ) = 0$ for $θ$ .

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

$θ = \arcsin (0)$

Simplify the right side.

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

$θ = 0$

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.

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

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.

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

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

Set $\sin (θ) - 1$ equal to $0$ .

$\sin (θ) - 1 = 0$

Solve $\sin (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\sin (θ) = 1$

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 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.

$θ = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

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

$θ = π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $π$ and $\frac{2}{2}$ .

$θ = \frac{π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

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 $1$ .

$2 π$

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

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

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

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

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

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

Solve for θ sin(theta)^2-sin(theta)=0

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