Trigonometry Examples

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

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

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

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $- 2 \cos (x) - 1 = 0$ .

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

Step 1.2.2

Add $1$ to both sides of the equation.

$- 2 \cos (x) = 1$

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $- 2$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

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

Step 1.2.3.2.1.2

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

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

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Move the negative in front of the fraction.

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

Step 1.2.4

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

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

Step 1.2.5

Simplify the right side.

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Step 1.2.5.1

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

$x = \frac{2 π}{3}$

Step 1.2.6

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 1.2.7

Simplify $2 π - \frac{2 π}{3}$ .

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Step 1.2.7.1

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 1.2.7.2

Combine fractions.

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Step 1.2.7.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 1.2.7.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 1.2.7.3

Simplify the numerator.

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Step 1.2.7.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 1.2.7.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 1.2.8

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

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Step 1.2.8.1

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

$\frac{2 π}{| b |}$

Step 1.2.8.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.8.3

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

$\frac{2 π}{1}$

Step 1.2.8.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.9

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 1.3

x-intercept(s) in point form.

x-intercept(s): $(\frac{2 π}{3} + 2 π n, 0), (\frac{4 π}{3} + 2 π n, 0)$ , for any integer $n$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

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

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

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

Step 2.2.2

Simplify $- 2 \cos (0) - 1$ .

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

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

$y = - 2 \cdot 1 - 1$

Step 2.2.2.1.2

Multiply $- 2$ by $1$ .

$y = - 2 - 1$

Step 2.2.2.2

Subtract $1$ from $- 2$ .

$y = - 3$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 3

List the intersections.

x-intercept(s): $(\frac{2 π}{3} + 2 π n, 0), (\frac{4 π}{3} + 2 π n, 0)$ , for any integer $n$

y-intercept(s): $(0, - 3)$

Step 4