Trigonometry Examples

$\tan^{2} (θ) + 5 \tan (θ) + 6 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

$u^{2} + 5 u + 6 = 0$

Step 1.2

Factor $u^{2} + 5 u + 6$ using the AC method.

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Step 1.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $5$ .

$2, 3$

Step 1.2.2

Write the factored form using these integers.

$(u + 2) (u + 3) = 0$

Step 1.3

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

$(\tan (θ) + 2) (\tan (θ) + 3) = 0$

Step 2

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

$\tan (θ) + 2 = 0$

$\tan (θ) + 3 = 0$

Step 3

Set $\tan (θ) + 2$ equal to $0$ and solve for $θ$ .

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Step 3.1

Set $\tan (θ) + 2$ equal to $0$ .

$\tan (θ) + 2 = 0$

Step 3.2

Solve $\tan (θ) + 2 = 0$ for $θ$ .

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Step 3.2.1

Subtract $2$ from both sides of the equation.

$\tan (θ) = - 2$

Step 3.2.2

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

$θ = \arctan (- 2)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Evaluate $\arctan (- 2)$ .

$θ = - 63.43494882$

Step 3.2.4

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 63.43494882 - 180$

Step 3.2.5

Simplify the expression to find the second solution.

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Step 3.2.5.1

Add $360 °$ to $- 63.43494882 - 180 °$ .

$θ = - 63.43494882 - 180 ° + 360 °$

Step 3.2.5.2

The resulting angle of $116.56505117 °$ is positive and coterminal with $- 63.43494882 - 180$ .

$θ = 116.56505117 °$

Step 3.2.6

Find the period of $\tan (θ)$ .

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Step 3.2.6.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 3.2.6.2

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

$\frac{180}{| 1 |}$

Step 3.2.6.3

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

$\frac{180}{1}$

Step 3.2.6.4

Divide $180$ by $1$ .

$180$

Step 3.2.7

Add $180$ to every negative angle to get positive angles.

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Step 3.2.7.1

Add $180$ to $- 63.43494882$ to find the positive angle.

$- 63.43494882 + 180$

Step 3.2.7.2

Subtract $63.43494882$ from $180$ .

$116.56505117$

Step 3.2.7.3

List the new angles.

$θ = 116.56505117$

Step 3.2.8

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 116.56505117 + 180 n, 116.56505117 + 180 n$ , for any integer $n$

Step 4

Set $\tan (θ) + 3$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $\tan (θ) + 3$ equal to $0$ .

$\tan (θ) + 3 = 0$

Step 4.2

Solve $\tan (θ) + 3 = 0$ for $θ$ .

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$\tan (θ) = - 3$

Step 4.2.2

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

$θ = \arctan (- 3)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Evaluate $\arctan (- 3)$ .

$θ = - 71.56505117$

Step 4.2.4

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 71.56505117 - 180$

Step 4.2.5

Simplify the expression to find the second solution.

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Step 4.2.5.1

Add $360 °$ to $- 71.56505117 - 180 °$ .

$θ = - 71.56505117 - 180 ° + 360 °$

Step 4.2.5.2

The resulting angle of $108.43494882 °$ is positive and coterminal with $- 71.56505117 - 180$ .

$θ = 108.43494882 °$

Step 4.2.6

Find the period of $\tan (θ)$ .

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Step 4.2.6.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 4.2.6.2

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

$\frac{180}{| 1 |}$

Step 4.2.6.3

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

$\frac{180}{1}$

Step 4.2.6.4

Divide $180$ by $1$ .

$180$

Step 4.2.7

Add $180$ to every negative angle to get positive angles.

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Step 4.2.7.1

Add $180$ to $- 71.56505117$ to find the positive angle.

$- 71.56505117 + 180$

Step 4.2.7.2

Subtract $71.56505117$ from $180$ .

$108.43494882$

Step 4.2.7.3

List the new angles.

$θ = 108.43494882$

Step 4.2.8

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 108.43494882 + 180 n, 108.43494882 + 180 n$ , for any integer $n$

Step 5

The final solution is all the values that make $(\tan (θ) + 2) (\tan (θ) + 3) = 0$ true.

$θ = 116.56505117 + 180 n, 108.43494882 + 180 n$ , for any integer $n$