Trigonometry Examples

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

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

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

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

Step 1.1.2

Apply the distributive property.

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

Step 1.1.3

Multiply $\cos (x)$ by $1$ .

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

Step 1.1.4

Rewrite using the commutative property of multiplication.

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

Step 1.1.5

Apply the sine double-angle identity.

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

Step 1.1.6

Multiply $- \sin (x) (2 \sin (x) \cos (x))$ .

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Step 1.1.6.1

Multiply $2$ by $- 1$ .

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

Step 1.1.6.2

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

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

Step 1.1.6.3

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

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

Step 1.1.6.4

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

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

Step 1.1.6.5

Add $1$ and $1$ .

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

Step 1.2

Simplify by adding terms.

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Step 1.2.1

Reorder the factors of $- 2 \cos (x) \sin^{2} (x)$ .

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

Step 1.2.2

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

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

Step 2

Factor $\cos (x) - 4 \sin^{2} (x) \cos (x)$ .

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Step 2.1

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

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Step 2.1.1

Raise $\cos (x)$ to the power of $1$ .

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.1.4

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

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

Step 2.2

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

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

Step 2.3

Rewrite $4 \sin^{2} (x)$ as ${(2 \sin (x))}^{2}$ .

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

Step 2.4

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 $b = 2 \sin (x)$ .

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

Step 2.5

Factor.

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Step 2.5.1

Multiply $2$ by $- 1$ .

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

Step 2.5.2

Remove unnecessary parentheses.

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

Step 3

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

$\cos (x) = 0$

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

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

Step 4

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

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Step 4.1

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

$\cos (x) = 0$

Step 4.2

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

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Step 4.2.1

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

$x = \arccos (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$x = \frac{π}{2}$

Step 4.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 4.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 4.2.4.1

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

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

Step 4.2.4.2

Combine fractions.

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Step 4.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 4.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 4.2.4.3

Simplify the numerator.

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Step 4.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 4.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 4.2.5

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

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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 $\cos (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

Subtract $1$ from both sides of the equation.

$2 \sin (x) = - 1$

Step 5.2.2

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

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Step 5.2.2.1

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

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

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.2.1.2

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

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

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

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

Step 5.2.3

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

$x = \arcsin (- \frac{1}{2})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$x = - \frac{π}{6}$

Step 5.2.5

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.

$x = 2 π + \frac{π}{6} + π$

Step 5.2.6

Simplify the expression to find the second solution.

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Step 5.2.6.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 5.2.6.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$x = \frac{7 π}{6}$

Step 5.2.7

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

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Step 5.2.7.1

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

$\frac{2 π}{| b |}$

Step 5.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.7.3

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

$\frac{2 π}{1}$

Step 5.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.8

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

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Step 5.2.8.1

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 5.2.8.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 5.2.8.3

Combine fractions.

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Step 5.2.8.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 5.2.8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 5.2.8.4

Simplify the numerator.

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Step 5.2.8.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 5.2.8.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 5.2.8.5

List the new angles.

$x = \frac{11 π}{6}$

Step 5.2.9

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

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 6

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

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Step 6.1

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

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

Step 6.2

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

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Step 6.2.1

Subtract $1$ from both sides of the equation.

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

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

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

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

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.

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

Step 6.2.3

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

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

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

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.

$x = π - \frac{π}{6}$

Step 6.2.6

Simplify $π - \frac{π}{6}$ .

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Step 6.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 6.2.6.2

Combine fractions.

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Step 6.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 6.2.6.3

Simplify the numerator.

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Step 6.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 6.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

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.

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

Step 7

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

$x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n, \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 8

Consolidate the answers.

$x = \frac{π}{6} + \frac{π n}{3}$ , for any integer $n$