Trigonometry Examples

$5 \cos (θ) = 5 \cos (2 θ)$

Step 1

Subtract $5 \cos (2 θ)$ from both sides of the equation.

$5 \cos (θ) - 5 \cos (2 θ) = 0$

Step 2

Simplify each term.

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Step 2.1

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

$5 \cos (θ) - 5 (2 \cos^{2} (θ) - 1) = 0$

Step 2.2

Apply the distributive property.

$5 \cos (θ) - 5 (2 \cos^{2} (θ)) - 5 \cdot - 1 = 0$

Step 2.3

Multiply $2$ by $- 5$ .

$5 \cos (θ) - 10 \cos^{2} (θ) - 5 \cdot - 1 = 0$

Step 2.4

Multiply $- 5$ by $- 1$ .

$5 \cos (θ) - 10 \cos^{2} (θ) + 5 = 0$

Step 3

Factor $5 \cos (θ) - 10 \cos^{2} (θ) + 5$ .

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Step 3.1

Factor $5$ out of $5 \cos (θ) - 10 \cos^{2} (θ) + 5$ .

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Step 3.1.1

Factor $5$ out of $5 \cos (θ)$ .

$5 \cos (θ) - 10 \cos^{2} (θ) + 5 = 0$

Step 3.1.2

Factor $5$ out of $- 10 \cos^{2} (θ)$ .

$5 \cos (θ) + 5 (- 2 \cos^{2} (θ)) + 5 = 0$

Step 3.1.3

Factor $5$ out of $5$ .

$5 \cos (θ) + 5 (- 2 \cos^{2} (θ)) + 5 (1) = 0$

Step 3.1.4

Factor $5$ out of $5 \cos (θ) + 5 (- 2 \cos^{2} (θ))$ .

$5 (\cos (θ) - 2 \cos^{2} (θ)) + 5 (1) = 0$

Step 3.1.5

Factor $5$ out of $5 (\cos (θ) - 2 \cos^{2} (θ)) + 5 (1)$ .

$5 (\cos (θ) - 2 \cos^{2} (θ) + 1) = 0$

Step 3.2

Let $u = \cos (θ)$ . Substitute $u$ for all occurrences of $\cos (θ)$ .

$5 (u - 2 u^{2} + 1) = 0$

Step 3.3

Factor by grouping.

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Step 3.3.1

Reorder terms.

$5 (- 2 u^{2} + u + 1) = 0$

Step 3.3.2

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 3.3.2.1

Multiply by $1$ .

$5 (- 2 u^{2} + 1 u + 1) = 0$

Step 3.3.2.2

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

$5 (- 2 u^{2} + (- 1 + 2) u + 1) = 0$

Step 3.3.2.3

Apply the distributive property.

$5 (- 2 u^{2} - 1 u + 2 u + 1) = 0$

Step 3.3.3

Factor out the greatest common factor from each group.

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Step 3.3.3.1

Group the first two terms and the last two terms.

$5 ((- 2 u^{2} - 1 u) + 2 u + 1) = 0$

Step 3.3.3.2

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

$5 (u (- 2 u - 1) - (- 2 u - 1)) = 0$

Step 3.3.4

Factor the polynomial by factoring out the greatest common factor, $- 2 u - 1$ .

$5 ((- 2 u - 1) (u - 1)) = 0$

Step 3.4

Factor.

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Step 3.4.1

Replace all occurrences of $u$ with $\cos (θ)$ .

$5 ((- 2 \cos (θ) - 1) (\cos (θ) - 1)) = 0$

Step 3.4.2

Remove unnecessary parentheses.

$5 (- 2 \cos (θ) - 1) (\cos (θ) - 1) = 0$

Step 4

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 5

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

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Step 5.1

Set $- 2 \cos (θ) - 1$ equal to $0$ .

$- 2 \cos (θ) - 1 = 0$

Step 5.2

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

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Step 5.2.1

Add $1$ to both sides of the equation.

$- 2 \cos (θ) = 1$

Step 5.2.2

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

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Step 5.2.2.1

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

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

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.

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

Step 5.2.2.2.1.2

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{- 2}$

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.

$\cos (θ) = - \frac{1}{2}$

Step 5.2.3

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

$θ = \arccos (- \frac{1}{2})$

Step 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 $\frac{2 π}{3}$ .

$θ = \frac{2 π}{3}$

Step 5.2.5

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.

$θ = 2 π - \frac{2 π}{3}$

Step 5.2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 5.2.6.1

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

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

Step 5.2.6.2

Combine fractions.

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Step 5.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 5.2.6.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Step 5.2.6.3

Simplify the numerator.

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Step 5.2.6.3.1

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Step 5.2.6.3.2

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

Step 5.2.7

Find the period of $\cos (θ)$ .

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

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

Step 6

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

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Step 6.1

Set $\cos (θ) - 1$ equal to $0$ .

$\cos (θ) - 1 = 0$

Step 6.2

Solve $\cos (θ) - 1 = 0$ for $θ$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$\cos (θ) = 1$

Step 6.2.2

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

$θ = \arccos (1)$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

The exact value of $\arccos (1)$ is $0$ .

$θ = 0$

Step 6.2.4

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.

$θ = 2 π - 0$

Step 6.2.5

Subtract $0$ from $2 π$ .

$θ = 2 π$

Step 6.2.6

Find the period of $\cos (θ)$ .

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Step 6.2.6.1

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

$\frac{2 π}{| b |}$

Step 6.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 6.2.6.3

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

$\frac{2 π}{1}$

Step 6.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 6.2.7

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

$θ = 2 π n, 2 π + 2 π n$ , for any integer $n$

Step 7

The final solution is all the values that make $5 (- 2 \cos (θ) - 1) (\cos (θ) - 1) = 0$ true.

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

Step 8

Consolidate the answers.

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