Trigonometry Examples

$\sin (2 θ) + \sin (4 θ) = 0$

Step 1

Simplify the left side of the equation.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the sine double-angle identity.

$2 \sin (θ) \cos (θ) + \sin (4 θ) = 0$

Step 1.1.2

Factor $2$ out of $4 θ$ .

$2 \sin (θ) \cos (θ) + \sin (2 (2 θ)) = 0$

Step 1.1.3

Apply the sine double-angle identity.

$2 \sin (θ) \cos (θ) + 2 (2 \sin (θ) \cos (θ)) \cos (2 θ) = 0$

Step 1.1.4

Multiply $2$ by $2$ .

$2 \sin (θ) \cos (θ) + 4 (\sin (θ) \cos (θ)) \cos (2 θ) = 0$

Step 1.1.5

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 1.1.6

Apply the distributive property.

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

Step 1.1.7

Multiply $4$ by $1$ .

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

Step 1.1.8

Multiply $\sin (θ)$ by $\sin^{2} (θ)$ by adding the exponents.

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Step 1.1.8.1

Move $\sin^{2} (θ)$ .

$2 \sin (θ) \cos (θ) + 4 \sin (θ) \cos (θ) + 4 (\sin^{2} (θ) \sin (θ)) \cos (θ) \cdot - 2 = 0$

Step 1.1.8.2

Multiply $\sin^{2} (θ)$ by $\sin (θ)$ .

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Step 1.1.8.2.1

Raise $\sin (θ)$ to the power of $1$ .

$2 \sin (θ) \cos (θ) + 4 \sin (θ) \cos (θ) + 4 (\sin^{2} (θ) \sin (θ)) \cos (θ) \cdot - 2 = 0$

Step 1.1.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.1.8.3

Add $2$ and $1$ .

$2 \sin (θ) \cos (θ) + 4 \sin (θ) \cos (θ) + 4 \sin^{3} (θ) \cos (θ) \cdot - 2 = 0$

Step 1.1.9

Multiply $- 2$ by $4$ .

$2 \sin (θ) \cos (θ) + 4 \sin (θ) \cos (θ) - 8 \sin^{3} (θ) \cos (θ) = 0$

Step 1.2

Add $2 \sin (θ) \cos (θ)$ and $4 \sin (θ) \cos (θ)$ .

$6 \sin (θ) \cos (θ) - 8 \sin^{3} (θ) \cos (θ) = 0$

Step 2

Factor $2 \sin (θ) \cos (θ)$ out of $6 \sin (θ) \cos (θ) - 8 \sin^{3} (θ) \cos (θ)$ .

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Step 2.1

Factor $2 \sin (θ) \cos (θ)$ out of $6 \sin (θ) \cos (θ)$ .

$2 \sin (θ) \cos (θ) \cdot 3 - 8 \sin^{3} (θ) \cos (θ) = 0$

Step 2.2

Factor $2 \sin (θ) \cos (θ)$ out of $- 8 \sin^{3} (θ) \cos (θ)$ .

$2 \sin (θ) \cos (θ) \cdot 3 + 2 \sin (θ) \cos (θ) (- 4 \sin^{2} (θ)) = 0$

Step 2.3

Factor $2 \sin (θ) \cos (θ)$ out of $2 \sin (θ) \cos (θ) \cdot 3 + 2 \sin (θ) \cos (θ) (- 4 \sin^{2} (θ))$ .

$2 \sin (θ) \cos (θ) (3 - 4 \sin^{2} (θ)) = 0$

Step 3

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

$\sin (θ) = 0$

$\cos (θ) = 0$

$3 - 4 \sin^{2} (θ) = 0$

Step 4

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

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Step 4.1

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Step 4.2

Solve $\sin (θ) = 0$ for $θ$ .

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Step 4.2.1

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

$θ = \arcsin (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$θ = 0$

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.

$θ = π - 0$

Step 4.2.4

Subtract $0$ from $π$ .

$θ = π$

Step 4.2.5

Find the period of $\sin (θ)$ .

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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 $\sin (θ)$ 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 $n$

Step 5

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

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Step 5.1

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Step 5.2

Solve $\cos (θ) = 0$ for $θ$ .

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Step 5.2.1

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

$θ = \arccos (0)$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

Step 5.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{2}$

Step 5.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 5.2.4.1

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

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

Step 5.2.4.2

Combine fractions.

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Step 5.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 5.2.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Step 5.2.4.3

Simplify the numerator.

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Step 5.2.4.3.1

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Step 5.2.4.3.2

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Step 5.2.5

Find the period of $\cos (θ)$ .

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Step 5.2.5.1

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

$\frac{2 π}{| b |}$

Step 5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.5.3

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

$\frac{2 π}{1}$

Step 5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.6

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

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

Step 6

Set $3 - 4 \sin^{2} (θ)$ equal to $0$ and solve for $θ$ .

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Step 6.1

Set $3 - 4 \sin^{2} (θ)$ equal to $0$ .

$3 - 4 \sin^{2} (θ) = 0$

Step 6.2

Solve $3 - 4 \sin^{2} (θ) = 0$ for $θ$ .

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Step 6.2.1

Subtract $3$ from both sides of the equation.

$- 4 \sin^{2} (θ) = - 3$

Step 6.2.2

Divide each term in $- 4 \sin^{2} (θ) = - 3$ by $- 4$ and simplify.

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Step 6.2.2.1

Divide each term in $- 4 \sin^{2} (θ) = - 3$ by $- 4$ .

$\frac{- 4 \sin^{2} (θ)}{- 4} = \frac{- 3}{- 4}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $- 4$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{- 4 \sin^{2} (θ)}{- 4} = \frac{- 3}{- 4}$

Step 6.2.2.2.1.2

Divide $\sin^{2} (θ)$ by $1$ .

$\sin^{2} (θ) = \frac{- 3}{- 4}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Dividing two negative values results in a positive value.

$\sin^{2} (θ) = \frac{3}{4}$

Step 6.2.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sin (θ) = \pm \sqrt{\frac{3}{4}}$

Step 6.2.4

Simplify $\pm \sqrt{\frac{3}{4}}$ .

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Step 6.2.4.1

Rewrite $\sqrt{\frac{3}{4}}$ as $\frac{\sqrt{3}}{\sqrt{4}}$ .

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{4}}$

Step 6.2.4.2

Simplify the denominator.

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Step 6.2.4.2.1

Rewrite $4$ as $2^{2}$ .

$\sin (θ) = \pm \frac{\sqrt{3}}{\sqrt{2^{2}}}$

Step 6.2.4.2.2

Pull terms out from under the radical, assuming positive real numbers.

$\sin (θ) = \pm \frac{\sqrt{3}}{2}$

Step 6.2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$\sin (θ) = \frac{\sqrt{3}}{2}$

Step 6.2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Step 6.2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sin (θ) = \frac{\sqrt{3}}{2}, - \frac{\sqrt{3}}{2}$

Step 6.2.6

Set up each of the solutions to solve for $θ$ .

$\sin (θ) = \frac{\sqrt{3}}{2}$

$\sin (θ) = - \frac{\sqrt{3}}{2}$

Step 6.2.7

Solve for $θ$ in $\sin (θ) = \frac{\sqrt{3}}{2}$ .

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Step 6.2.7.1

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

$θ = \arcsin (\frac{\sqrt{3}}{2})$

Step 6.2.7.2

Simplify the right side.

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Step 6.2.7.2.1

The exact value of $\arcsin (\frac{\sqrt{3}}{2})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 6.2.7.3

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.

$θ = π - \frac{π}{3}$

Step 6.2.7.4

Simplify $π - \frac{π}{3}$ .

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Step 6.2.7.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = π \cdot \frac{3}{3} - \frac{π}{3}$

Step 6.2.7.4.2

Combine fractions.

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Step 6.2.7.4.2.1

Combine $π$ and $\frac{3}{3}$ .

$θ = \frac{π \cdot 3}{3} - \frac{π}{3}$

Step 6.2.7.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 - π}{3}$

Step 6.2.7.4.3

Simplify the numerator.

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Step 6.2.7.4.3.1

Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π - π}{3}$

Step 6.2.7.4.3.2

Subtract $π$ from $3 π$ .

$θ = \frac{2 π}{3}$

Step 6.2.7.5

Find the period of $\sin (θ)$ .

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Step 6.2.7.5.1

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

$\frac{2 π}{| b |}$

Step 6.2.7.5.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.7.5.3

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

$\frac{2 π}{1}$

Step 6.2.7.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.7.6

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

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

Step 6.2.8

Solve for $θ$ in $\sin (θ) = - \frac{\sqrt{3}}{2}$ .

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Step 6.2.8.1

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

$θ = \arcsin (- \frac{\sqrt{3}}{2})$

Step 6.2.8.2

Simplify the right side.

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Step 6.2.8.2.1

The exact value of $\arcsin (- \frac{\sqrt{3}}{2})$ is $- \frac{π}{3}$ .

$θ = - \frac{π}{3}$

Step 6.2.8.3

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.

$θ = 2 π + \frac{π}{3} + π$

Step 6.2.8.4

Simplify the expression to find the second solution.

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Step 6.2.8.4.1

Subtract $2 π$ from $2 π + \frac{π}{3} + π$ .

$θ = 2 π + \frac{π}{3} + π - 2 π$

Step 6.2.8.4.2

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$θ = \frac{4 π}{3}$

Step 6.2.8.5

Find the period of $\sin (θ)$ .

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Step 6.2.8.5.1

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

$\frac{2 π}{| b |}$

Step 6.2.8.5.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.8.5.3

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

$\frac{2 π}{1}$

Step 6.2.8.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.8.6

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

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Step 6.2.8.6.1

Add $2 π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + 2 π$

Step 6.2.8.6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 6.2.8.6.3

Combine fractions.

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Step 6.2.8.6.3.1

Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 6.2.8.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Step 6.2.8.6.4

Simplify the numerator.

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Step 6.2.8.6.4.1

Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Step 6.2.8.6.4.2

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

Step 6.2.8.6.5

List the new angles.

$θ = \frac{5 π}{3}$

Step 6.2.8.7

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

$θ = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 6.2.9

List all of the solutions.

$θ = \frac{π}{3} + 2 π n, \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 6.2.10

Consolidate the solutions.

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Step 6.2.10.1

Consolidate $\frac{π}{3} + 2 π n$ and $\frac{4 π}{3} + 2 π n$ to $\frac{π}{3} + π n$ .

$θ = \frac{π}{3} + π n, \frac{2 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 6.2.10.2

Consolidate $\frac{2 π}{3} + 2 π n$ and $\frac{5 π}{3} + 2 π n$ to $\frac{2 π}{3} + π n$ .

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

Step 7

The final solution is all the values that make $2 \sin (θ) \cos (θ) (3 - 4 \sin^{2} (θ)) = 0$ true.

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

Step 8

Consolidate the answers.

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Step 8.1

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

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

Step 8.2

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

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

Step 8.3

Consolidate $π n$ and $\frac{π}{2} + π n$ to $\frac{π n}{2}$ .

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