Trigonometry Examples

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

Step 1

Simplify the left side of the equation.

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Step 1.1

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

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

Step 1.2

Add $2 \cos^{2} (x)$ and $2 \cos^{2} (x)$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

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 $a = 2 \cos (x)$ and $b = 1$ .

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

Step 3

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

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

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$2 \cos (x) = - 1$

Step 4.2.2

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

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Step 4.2.2.1

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

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

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

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

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Move the negative in front of the fraction.

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

Step 4.2.3

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

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

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$x = \frac{2 π}{3}$

Step 4.2.5

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

$x = 2 π - \frac{2 π}{3}$

Step 4.2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 4.2.6.1

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

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 4.2.6.2

Combine fractions.

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Step 4.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 4.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 4.2.6.3

Simplify the numerator.

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Step 4.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 4.2.6.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 4.2.7

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

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Step 4.2.7.1

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

$\frac{2 π}{| b |}$

Step 4.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.7.3

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

$\frac{2 π}{1}$

Step 4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.8

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 5.2.2

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

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Step 5.2.2.1

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

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

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.

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

Step 5.2.2.2.1.2

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

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

Step 5.2.3

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

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

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 5.2.5

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.

$x = 2 π - \frac{π}{3}$

Step 5.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 5.2.6.1

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

$x = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 5.2.6.2

Combine fractions.

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Step 5.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 5.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 5.2.6.3

Simplify the numerator.

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Step 5.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 5.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 5.2.7

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

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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

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

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

Step 6

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

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

Step 7

Consolidate the answers.

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Step 7.1

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

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

Step 7.2

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

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