Trigonometry Examples

$\cos (2 θ) + \cos (θ) = 0$ , $0 \leq θ \leq 360$

Step 1

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

$2 \cos^{2} (θ) - 1 + \cos (θ) = 0$

Step 2

Factor by grouping.

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Step 2.1

Reorder terms.

$2 \cos^{2} (θ) + \cos (θ) - 1 = 0$

Step 2.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 2.2.1

Multiply by $1$ .

$2 \cos^{2} (θ) + 1 \cos (θ) - 1 = 0$

Step 2.2.2

Rewrite $1$ as $- 1$ plus $2$

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

Step 2.2.3

Apply the distributive property.

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

Step 2.3

Factor out the greatest common factor from each group.

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Step 2.3.1

Group the first two terms and the last two terms.

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

Step 2.3.2

Factor out the greatest common factor (GCF) from each group.

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

Step 2.4

Factor the polynomial by factoring out the greatest common factor, $2 \cos (θ) - 1$ .

$(2 \cos (θ) - 1) (\cos (θ) + 1) = 0$

Step 3

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

$2 \cos (θ) - 1 = 0$

$\cos (θ) + 1 = 0$

Step 4

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

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Step 4.1

Set $2 \cos (θ) - 1$ equal to $0$ .

$2 \cos (θ) - 1 = 0$

Step 4.2

Solve $2 \cos (θ) - 1 = 0$ for $θ$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$2 \cos (θ) = 1$

Step 4.2.2

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

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Step 4.2.2.1

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

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

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.

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

Step 4.2.2.2.1.2

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{2}$

Step 4.2.3

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

$θ = \arccos (\frac{1}{2})$

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 $60$ .

$θ = 60$

Step 4.2.5

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 60$

Step 4.2.6

Subtract $60$ from $360$ .

$θ = 300$

Step 4.2.7

Find the period of $\cos (θ)$ .

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Step 4.2.7.1

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

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

Step 4.2.7.2

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

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

Step 4.2.7.3

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

$\frac{360}{1}$

Step 4.2.7.4

Divide $360$ by $1$ .

$360$

Step 4.2.8

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

$θ = 60 + 360 n, 300 + 360 n$ , for any integer $n$

Step 5

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

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Step 5.1

Set $\cos (θ) + 1$ equal to $0$ .

$\cos (θ) + 1 = 0$

Step 5.2

Solve $\cos (θ) + 1 = 0$ for $θ$ .

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Step 5.2.1

Subtract $1$ from both sides of the equation.

$\cos (θ) = - 1$

Step 5.2.2

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

$θ = \arccos (- 1)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

The exact value of $\arccos (- 1)$ is $180$ .

$θ = 180$

Step 5.2.4

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the third quadrant.

$θ = 360 - 180$

Step 5.2.5

Subtract $180$ from $360$ .

$θ = 180$

Step 5.2.6

Find the period of $\cos (θ)$ .

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Step 5.2.6.1

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

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

Step 5.2.6.2

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

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

Step 5.2.6.3

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

$\frac{360}{1}$

Step 5.2.6.4

Divide $360$ by $1$ .

$360$

Step 5.2.7

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

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

Step 6

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

$θ = 60 + 360 n, 300 + 360 n, 180 + 360 n$ , for any integer $n$

Step 7

Consolidate the answers.

$θ = 60 + 120 n$ , for any integer $n$

Step 8

Find the values of $n$ that produce a value within the interval $[0, 360]$ .

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Step 8.1

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 360]$ .

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Step 8.1.1

Plug in $0$ for $n$ .

$60 + 120 (0)$

Step 8.1.2

Simplify.

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Step 8.1.2.1

Multiply $120$ by $0$ .

$60 + 0$

Step 8.1.2.2

Add $60$ and $0$ .

$60$

Step 8.1.3

The interval $[0, 360]$ contains $60$ .

$θ = 60$

Step 8.2

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 360]$ .

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Step 8.2.1

Plug in $1$ for $n$ .

$60 + 120 (1)$

Step 8.2.2

Simplify.

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Step 8.2.2.1

Multiply $120$ by $1$ .

$60 + 120$

Step 8.2.2.2

Add $60$ and $120$ .

$180$

Step 8.2.3

The interval $[0, 360]$ contains $180$ .

$θ = 60, 180$

Step 8.3

Plug in $2$ for $n$ and simplify to see if the solution is contained in $[0, 360]$ .

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Step 8.3.1

Plug in $2$ for $n$ .

$60 + 120 (2)$

Step 8.3.2

Simplify.

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Step 8.3.2.1

Multiply $120$ by $2$ .

$60 + 240$

Step 8.3.2.2

Add $60$ and $240$ .

$300$

Step 8.3.3

The interval $[0, 360]$ contains $300$ .

$θ = 60, 180, 300$