Trigonometry Examples

$\tan^{2} (θ) - 6 \tan (θ) + 5 = 0$

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 $\tan (θ)$ .

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

Step 1.2

Factor $u^{2} - 6 u + 5$ 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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Step 1.2.2

Write the factored form using these integers.

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

Step 1.3

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

$(\tan (θ) - 5) (\tan (θ) - 1) = 0$

Step 2

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

$\tan (θ) - 5 = 0$

$\tan (θ) - 1 = 0$

Step 3

Set $\tan (θ) - 5$ equal to $0$ and solve for $θ$ .

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Step 3.1

Set $\tan (θ) - 5$ equal to $0$ .

$\tan (θ) - 5 = 0$

Step 3.2

Solve $\tan (θ) - 5 = 0$ for $θ$ .

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Step 3.2.1

Add $5$ to both sides of the equation.

$\tan (θ) = 5$

Step 3.2.2

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

$θ = \arctan (5)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Evaluate $\arctan (5)$ .

$θ = 78.69006752$

Step 3.2.4

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

$θ = 180 + 78.69006752$

Step 3.2.5

Add $180$ and $78.69006752$ .

$θ = 258.69006752$

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

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

$θ = 78.69006752 + 180 n, 258.69006752 + 180 n$ , for any integer $n$

Step 4

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

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Step 4.1

Set $\tan (θ) - 1$ equal to $0$ .

$\tan (θ) - 1 = 0$

Step 4.2

Solve $\tan (θ) - 1 = 0$ for $θ$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$\tan (θ) = 1$

Step 4.2.2

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

$θ = \arctan (1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of $\arctan (1)$ is $45$ .

$θ = 45$

Step 4.2.4

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

$θ = 180 + 45$

Step 4.2.5

Add $180$ and $45$ .

$θ = 225$

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

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

$θ = 45 + 180 n, 225 + 180 n$ , for any integer $n$

Step 5

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

$θ = 78.69006752 + 180 n, 258.69006752 + 180 n, 45 + 180 n, 225 + 180 n$ , for any integer $n$

Step 6

Consolidate the answers.

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Step 6.1

Consolidate $78.69006752 + 180 n$ and $258.69006752 + 180 n$ to $78.69006752 + 180 n$ .

$θ = 78.69006752 + 180 n, 45 + 180 n, 225 + 180 n$ , for any integer $n$

Step 6.2

Consolidate $45 + 180 n$ and $225 + 180 n$ to $45 + 180 n$ .

$θ = 78.69006752 + 180 n, 45 + 180 n$ , for any integer $n$