Trigonometry Examples

$\sin (3 x) + \sin (x) = 0$

Step 1

Simplify the left side of the equation.

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Step 1.1

Apply the sine triple-angle identity.

$- 4 \sin^{3} (x) + 3 \sin (x) + \sin (x) = 0$

Step 1.2

Add $3 \sin (x)$ and $\sin (x)$ .

$- 4 \sin^{3} (x) + 4 \sin (x) = 0$

Step 2

Factor $- 4 \sin^{3} (x) + 4 \sin (x)$ .

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Step 2.1

Factor $4 \sin (x)$ out of $- 4 \sin^{3} (x) + 4 \sin (x)$ .

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Step 2.1.1

Factor $4 \sin (x)$ out of $- 4 \sin^{3} (x)$ .

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

Step 2.1.2

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

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

Step 2.1.3

Factor $4 \sin (x)$ out of $4 \sin (x) (- \sin^{2} (x)) + 4 \sin (x) \cdot 1$ .

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

Step 2.2

Rewrite $1$ as $1^{2}$ .

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

Step 2.3

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

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

Step 2.4

Factor.

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Step 2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \sin (x)$ .

$4 \sin (x) ((1 + \sin (x)) (1 - \sin (x))) = 0$

Step 2.4.2

Remove unnecessary parentheses.

$4 \sin (x) (1 + \sin (x)) (1 - \sin (x)) = 0$

Step 3

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

$\sin (x) = 0$

$1 + \sin (x) = 0$

$1 - \sin (x) = 0$

Step 4

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

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Step 4.1

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

$\sin (x) = 0$

Step 4.2

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

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Step 4.2.1

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

$x = \arcsin (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$x = 0$

Step 4.2.3

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.

$x = π - 0$

Step 4.2.4

Subtract $0$ from $π$ .

$x = π$

Step 4.2.5

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

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Step 4.2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.5.3

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

$\frac{2 π}{1}$

Step 4.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.6

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 5

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

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Step 5.1

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

$1 + \sin (x) = 0$

Step 5.2

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

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Step 5.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 5.2.2

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

$x = \arcsin (- 1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

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

$x = - \frac{π}{2}$

Step 5.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 5.2.5

Simplify the expression to find the second solution.

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Step 5.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 5.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 5.2.6

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

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Step 5.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 5.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

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

$\frac{2 π}{1}$

Step 5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.7

Add $2 π$ to every negative angle to get positive angles.

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Step 5.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 5.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 5.2.7.3

Combine fractions.

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Step 5.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 5.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 5.2.7.4

Simplify the numerator.

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Step 5.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 5.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 5.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 5.2.8

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

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 6

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

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Step 6.1

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

$1 - \sin (x) = 0$

Step 6.2

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

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$- \sin (x) = - 1$

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 6.2.2.2.2

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

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

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 1$ by $- 1$ .

$\sin (x) = 1$

Step 6.2.3

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

$x = \arcsin (1)$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

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

$x = \frac{π}{2}$

Step 6.2.5

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.

$x = π - \frac{π}{2}$

Step 6.2.6

Simplify $π - \frac{π}{2}$ .

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Step 6.2.6.1

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

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

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 6.2.6.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 6.2.7

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

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Step 6.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.7.3

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

$\frac{2 π}{1}$

Step 6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8

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

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

Step 7

The final solution is all the values that make $4 \sin (x) (1 + \sin (x)) (1 - \sin (x)) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{2} + 2 π n$ , for any integer $n$

Step 8

Consolidate the answers.

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