Solve for x in Degrees sin(x)^2+sin(x)=0

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Trigonometry

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

Factor the left side of the equation.

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

$u^{2} + u = 0$

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

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

$u \cdot u + u = 0$

Raise $u$ to the power of $1$ .

$u \cdot u + u = 0$

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

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

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

$u (u + 1) = 0$

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

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

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

$\sin (x) = 0$

$\sin (x) + 1 = 0$

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

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

$\sin (x) = 0$

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

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

$x = \arcsin (0)$

Simplify the right side.

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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 $180$ to find the solution in the second quadrant.

$x = 180 - 0$

Subtract $0$ from $180$ .

$x = 180$

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

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 360 n, 180 + 360 n$ , for any integer $n$

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

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

$\sin (x) + 1 = 0$

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

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Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

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

$x = \arcsin (- 1)$

Simplify the right side.

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The exact value of $\arcsin (- 1)$ is $- 90$ .

$x = - 90$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$x = 360 + 90 + 180$

Simplify the expression to find the second solution.

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Subtract $360 °$ from $360 + 90 + 180 °$ .

$x = 360 + 90 + 180 ° - 360 °$

The resulting angle of $270 °$ is positive, less than $360 °$ , and coterminal with $360 + 90 + 180$ .

$x = 270 °$

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

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

Add $360$ to every negative angle to get positive angles.

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Add $360$ to $- 90$ to find the positive angle.

$- 90 + 360$

Subtract $90$ from $360$ .

$270$

List the new angles.

$x = 270$

The period of the $\sin (x)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$x = 270 + 360 n, 270 + 360 n$ , for any integer $n$

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

$x = 360 n, 180 + 360 n, 270 + 360 n$ , for any integer $n$

Consolidate $360 n$ and $180 + 360 n$ to $180 n$ .

$x = 180 n, 270 + 360 n$ , for any integer $n$

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