Trigonometry Examples

$6 \sec^{2} (θ) \tan (θ) = 12 \tan (θ)$

Step 1

Subtract $12 \tan (θ)$ from both sides of the equation.

$6 \sec^{2} (θ) \tan (θ) - 12 \tan (θ) = 0$

Step 2

Factor $6 \tan (θ)$ out of $6 \sec^{2} (θ) \tan (θ) - 12 \tan (θ)$ .

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Step 2.1

Factor $6 \tan (θ)$ out of $6 \sec^{2} (θ) \tan (θ)$ .

$6 \tan (θ) \sec^{2} (θ) - 12 \tan (θ) = 0$

Step 2.2

Factor $6 \tan (θ)$ out of $- 12 \tan (θ)$ .

$6 \tan (θ) \sec^{2} (θ) + 6 \tan (θ) \cdot - 2 = 0$

Step 2.3

Factor $6 \tan (θ)$ out of $6 \tan (θ) \sec^{2} (θ) + 6 \tan (θ) \cdot - 2$ .

$6 \tan (θ) (\sec^{2} (θ) - 2) = 0$

Step 3

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

$\tan (θ) = 0$

$\sec^{2} (θ) - 2 = 0$

Step 4

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

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Step 4.1

Set $\tan (θ)$ equal to $0$ .

$\tan (θ) = 0$

Step 4.2

Solve $\tan (θ) = 0$ for $θ$ .

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Step 4.2.1

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

$θ = \arctan (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

The exact value of $\arctan (0)$ is $0$ .

$θ = 0$

Step 4.2.3

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.

$θ = 180 + 0$

Step 4.2.4

Add $180$ and $0$ .

$θ = 180$

Step 4.2.5

Find the period of $\tan (θ)$ .

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Step 4.2.5.1

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

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

Step 4.2.5.2

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

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

Step 4.2.5.3

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

$\frac{180}{1}$

Step 4.2.5.4

Divide $180$ by $1$ .

$180$

Step 4.2.6

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

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

Step 5

Set $\sec^{2} (θ) - 2$ equal to $0$ and solve for $θ$ .

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Step 5.1

Set $\sec^{2} (θ) - 2$ equal to $0$ .

$\sec^{2} (θ) - 2 = 0$

Step 5.2

Solve $\sec^{2} (θ) - 2 = 0$ for $θ$ .

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Step 5.2.1

Add $2$ to both sides of the equation.

$\sec^{2} (θ) = 2$

Step 5.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sec (θ) = \pm \sqrt{2}$

Step 5.2.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.2.3.1

First, use the positive value of the $\pm$ to find the first solution.

$\sec (θ) = \sqrt{2}$

Step 5.2.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sec (θ) = - \sqrt{2}$

Step 5.2.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sec (θ) = \sqrt{2}, - \sqrt{2}$

Step 5.2.4

Set up each of the solutions to solve for $θ$ .

$\sec (θ) = \sqrt{2}$

$\sec (θ) = - \sqrt{2}$

Step 5.2.5

Solve for $θ$ in $\sec (θ) = \sqrt{2}$ .

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Step 5.2.5.1

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

$θ = arcsec (\sqrt{2})$

Step 5.2.5.2

Simplify the right side.

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Step 5.2.5.2.1

The exact value of $arcsec (\sqrt{2})$ is $45$ .

$θ = 45$

Step 5.2.5.3

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.

$θ = 360 - 45$

Step 5.2.5.4

Subtract $45$ from $360$ .

$θ = 315$

Step 5.2.5.5

Find the period of $\sec (θ)$ .

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Step 5.2.5.5.1

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

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

Step 5.2.5.5.2

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

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

Step 5.2.5.5.3

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

$\frac{360}{1}$

Step 5.2.5.5.4

Divide $360$ by $1$ .

$360$

Step 5.2.5.6

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

$θ = 45 + 360 n, 315 + 360 n$ , for any integer $n$

Step 5.2.6

Solve for $θ$ in $\sec (θ) = - \sqrt{2}$ .

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Step 5.2.6.1

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

$θ = arcsec (- \sqrt{2})$

Step 5.2.6.2

Simplify the right side.

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Step 5.2.6.2.1

The exact value of $arcsec (- \sqrt{2})$ is $135$ .

$θ = 135$

Step 5.2.6.3

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

$θ = 360 - 135$

Step 5.2.6.4

Subtract $135$ from $360$ .

$θ = 225$

Step 5.2.6.5

Find the period of $\sec (θ)$ .

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Step 5.2.6.5.1

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

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

Step 5.2.6.5.2

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

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

Step 5.2.6.5.3

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

$\frac{360}{1}$

Step 5.2.6.5.4

Divide $360$ by $1$ .

$360$

Step 5.2.6.6

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

$θ = 135 + 360 n, 225 + 360 n$ , for any integer $n$

Step 5.2.7

List all of the solutions.

$θ = 45 + 360 n, 315 + 360 n, 135 + 360 n, 225 + 360 n$ , for any integer $n$

Step 5.2.8

Consolidate the answers.

$θ = 45 + 90 n$ , for any integer $n$

Step 6

The final solution is all the values that make $6 \tan (θ) (\sec^{2} (θ) - 2) = 0$ true.

$θ = 180 n, 180 + 180 n, 45 + 90 n$ , for any integer $n$

Step 7

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

$θ = 180 n, 45 + 90 n$ , for any integer $n$