Trigonometry Examples

$\sec^{2} (θ) - 9 \sec (θ) + 20 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

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

$u^{2} - 9 u + 20 = 0$

Step 1.2

Factor $u^{2} - 9 u + 20$ 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 $20$ and whose sum is $- 9$ .

$- 5, - 4$

Step 1.2.2

Write the factored form using these integers.

$(u - 5) (u - 4) = 0$

Step 1.3

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

$(\sec (θ) - 5) (\sec (θ) - 4) = 0$

Step 2

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

$\sec (θ) - 5 = 0$

$\sec (θ) - 4 = 0$

Step 3

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

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Step 3.1

Set $\sec (θ) - 5$ equal to $0$ .

$\sec (θ) - 5 = 0$

Step 3.2

Solve $\sec (θ) - 5 = 0$ for $θ$ .

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Step 3.2.1

Add $5$ to both sides of the equation.

$\sec (θ) = 5$

Step 3.2.2

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

$θ = arcsec (5)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Evaluate $arcsec (5)$ .

$θ = 78.46304096$

Step 3.2.4

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 78.46304096$

Step 3.2.5

Subtract $78.46304096$ from $360$ .

$θ = 281.53695903$

Step 3.2.6

Find the period of $\sec (θ)$ .

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Step 3.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 3.2.6.2

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

$\frac{360}{| 1 |}$

Step 3.2.6.3

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

$\frac{360}{1}$

Step 3.2.6.4

Divide $360$ by $1$ .

$360$

Step 3.2.7

The period of the $\sec (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 78.46304096 + 360 n, 281.53695903 + 360 n$ , for any integer $n$

Step 4

Set $\sec (θ) - 4$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $\sec (θ) - 4$ equal to $0$ .

$\sec (θ) - 4 = 0$

Step 4.2

Solve $\sec (θ) - 4 = 0$ for $θ$ .

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Step 4.2.1

Add $4$ to both sides of the equation.

$\sec (θ) = 4$

Step 4.2.2

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

$θ = arcsec (4)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Evaluate $arcsec (4)$ .

$θ = 75.52248781$

Step 4.2.4

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $360$ to find the solution in the fourth quadrant.

$θ = 360 - 75.52248781$

Step 4.2.5

Subtract $75.52248781$ from $360$ .

$θ = 284.47751218$

Step 4.2.6

Find the period of $\sec (θ)$ .

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Step 4.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 4.2.6.2

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

$\frac{360}{| 1 |}$

Step 4.2.6.3

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

$\frac{360}{1}$

Step 4.2.6.4

Divide $360$ by $1$ .

$360$

Step 4.2.7

The period of the $\sec (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 75.52248781 + 360 n, 284.47751218 + 360 n$ , for any integer $n$

Step 5

The final solution is all the values that make $(\sec (θ) - 5) (\sec (θ) - 4) = 0$ true.

$θ = 78.46304096 + 360 n, 281.53695903 + 360 n, 75.52248781 + 360 n, 284.47751218 + 360 n$ , for any integer $n$