Trigonometry Examples

$\frac{- 3 \sin (x) + 4 \cos (x)}{5 \cos (x) + 2 \sin (x)}$

Step 1

Set the denominator in $\frac{- 3 \sin (x) + 4 \cos (x)}{5 \cos (x) + 2 \sin (x)}$ equal to $0$ to find where the expression is undefined.

$5 \cos (x) + 2 \sin (x) = 0$

Step 2

Solve for $x$ .

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Step 2.1

Divide each term in the equation by $\cos (x)$ .

$\frac{5 \cos (x)}{\cos (x)} + \frac{2 \sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 2.2

Cancel the common factor of $\cos (x)$ .

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Step 2.2.1

Cancel the common factor.

$\frac{5 \cos (x)}{\cos (x)} + \frac{2 \sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 2.2.2

Divide $5$ by $1$ .

$5 + \frac{2 \sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 2.3

Separate fractions.

$5 + \frac{2}{1} \cdot \frac{\sin (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 2.4

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$5 + \frac{2}{1} \cdot \tan (x) = \frac{0}{\cos (x)}$

Step 2.5

Divide $2$ by $1$ .

$5 + 2 \tan (x) = \frac{0}{\cos (x)}$

Step 2.6

Separate fractions.

$5 + 2 \tan (x) = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 2.7

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$5 + 2 \tan (x) = \frac{0}{1} \cdot \sec (x)$

Step 2.8

Divide $0$ by $1$ .

$5 + 2 \tan (x) = 0 \sec (x)$

Step 2.9

Multiply $0$ by $\sec (x)$ .

$5 + 2 \tan (x) = 0$

Step 2.10

Subtract $5$ from both sides of the equation.

$2 \tan (x) = - 5$

Step 2.11

Divide each term in $2 \tan (x) = - 5$ by $2$ and simplify.

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Step 2.11.1

Divide each term in $2 \tan (x) = - 5$ by $2$ .

$\frac{2 \tan (x)}{2} = \frac{- 5}{2}$

Step 2.11.2

Simplify the left side.

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Step 2.11.2.1

Cancel the common factor of $2$ .

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Step 2.11.2.1.1

Cancel the common factor.

$\frac{2 \tan (x)}{2} = \frac{- 5}{2}$

Step 2.11.2.1.2

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

$\tan (x) = \frac{- 5}{2}$

Step 2.11.3

Simplify the right side.

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Step 2.11.3.1

Move the negative in front of the fraction.

$\tan (x) = - \frac{5}{2}$

Step 2.12

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

$x = \arctan (- \frac{5}{2})$

Step 2.13

Simplify the right side.

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Step 2.13.1

Evaluate $\arctan (- \frac{5}{2})$ .

$x = - 1.19028994$

Step 2.14

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

$x = - 1.19028994 - (3.14159265)$

Step 2.15

Simplify the expression to find the second solution.

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Step 2.15.1

Add $2 π$ to $- 1.19028994 - (3.14159265)$ .

$x = - 1.19028994 - (3.14159265) + 2 π$

Step 2.15.2

The resulting angle of $1.9513027$ is positive and coterminal with $- 1.19028994 - (3.14159265)$ .

$x = 1.9513027$

Step 2.16

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

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Step 2.16.1

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

$\frac{π}{| b |}$

Step 2.16.2

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

$\frac{π}{| 1 |}$

Step 2.16.3

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

$\frac{π}{1}$

Step 2.16.4

Divide $π$ by $1$ .

$π$

Step 2.17

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

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Step 2.17.1

Add $π$ to $- 1.19028994$ to find the positive angle.

$- 1.19028994 + π$

Step 2.17.2

Replace with decimal approximation.

$3.14159265 - 1.19028994$

Step 2.17.3

Subtract $1.19028994$ from $3.14159265$ .

$1.9513027$

Step 2.17.4

List the new angles.

$x = 1.9513027$

Step 2.18

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

$x = 1.9513027 + π n, 1.9513027 + π n$ , for any integer $n$

Step 3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

${x | x = 1.9513027 + π n, x = 1.9513027 + π n}$ $n$ , for any integer $n$

Step 4