Algebra Examples

$\tan^{2} (θ) = \frac{3}{2} \sec (θ)$

Step 1

Multiply each term by a factor of $1$ that will equate all the denominators. In this case, all terms need a denominator of $2$ .

$\tan^{2} (θ) \cdot \frac{2}{2} = \frac{3}{2} \cdot \sec (θ) \cdot \frac{2}{2}$

Step 2

Multiply the expression by a factor of $1$ to create the least common denominator (LCD) of $2$ .

$\tan^{2} (θ) \cdot 2$

Step 3

Move $2$ to the left of $\tan^{2} (θ)$ .

$\frac{2 \tan^{2} (θ)}{2} = \frac{3}{2} \cdot \sec (θ) \cdot \frac{2}{2}$

Step 4

Multiply the expression by a factor of $1$ to create the least common denominator (LCD) of $2$ .

$\sec (θ) \cdot 2$

Step 5

Move $2$ to the left of $\sec (θ)$ .

$\frac{2 \tan^{2} (θ)}{2} = \frac{3}{2} \cdot \frac{2 \sec (θ)}{2}$

Step 6

Replace the $\tan^{2} (θ)$ with $\sec^{2} (θ) - 1$ based on the $\tan^{2} (x) + 1 = \sec^{2} (x)$ identity.

$(\sec^{2} (θ) - 1) \cdot \frac{2}{2} = \frac{3}{2} \cdot \sec (θ) \cdot \frac{2}{2}$

Step 7

Divide $2$ by $2$ .

$(\sec^{2} (θ) - 1) \cdot 1 = \frac{3}{2} \cdot \sec (θ) \cdot \frac{2}{2}$

Step 8

Multiply $\sec^{2} (θ) - 1$ by $1$ .

$\sec^{2} (θ) - 1 = \frac{3}{2} \cdot \sec (θ) \cdot \frac{2}{2}$

Step 9

Substitute $u$ for $\sec (θ)$ .

${(u)}^{2} - 1 = \frac{3}{2} (u) \cdot \frac{2}{2}$

Step 10

Simplify $\frac{3}{2} u \cdot \frac{2}{2}$ .

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Step 10.1

Rewrite.

$u^{2} - 1 = 0 + 0 + \frac{3}{2} u \cdot \frac{2}{2}$

Step 10.2

Simplify by adding zeros.

$u^{2} - 1 = \frac{3}{2} u \cdot \frac{2}{2}$

Step 10.3

Combine $\frac{3}{2}$ and $u$ .

$u^{2} - 1 = \frac{3 u}{2} \cdot \frac{2}{2}$

Step 10.4

Cancel the common factor of $2$ .

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Step 10.4.1

Cancel the common factor.

$u^{2} - 1 = \frac{3 u}{2} \cdot \frac{2}{2}$

Step 10.4.2

Rewrite the expression.

$u^{2} - 1 = \frac{3 u}{2} \cdot 1$

Step 10.5

Multiply $\frac{3 u}{2}$ by $1$ .

$u^{2} - 1 = \frac{3 u}{2}$

Step 11

Subtract $\frac{3 u}{2}$ from both sides of the equation.

$u^{2} - 1 - \frac{3 u}{2} = 0$

Step 12

Multiply through by the least common denominator $2$ , then simplify.

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Step 12.1

Apply the distributive property.

$2 u^{2} + 2 \cdot - 1 + 2 (- \frac{3 u}{2}) = 0$

Step 12.2

Simplify.

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Step 12.2.1

Multiply $2$ by $- 1$ .

$2 u^{2} - 2 + 2 (- \frac{3 u}{2}) = 0$

Step 12.2.2

Cancel the common factor of $2$ .

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Step 12.2.2.1

Move the leading negative in $- \frac{3 u}{2}$ into the numerator.

$2 u^{2} - 2 + 2 \frac{- 3 u}{2} = 0$

Step 12.2.2.2

Cancel the common factor.

$2 u^{2} - 2 + 2 \frac{- 3 u}{2} = 0$

Step 12.2.2.3

Rewrite the expression.

$2 u^{2} - 2 - 3 u = 0$

Step 12.3

Move $- 2$ .

$2 u^{2} - 3 u - 2 = 0$

Step 13

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 14

Substitute the values $a = 2$ , $b = - 3$ , and $c = - 2$ into the quadratic formula and solve for $u$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 2)}}{2 \cdot 2}$

Step 15

Simplify.

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Step 15.1

Simplify the numerator.

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Step 15.1.1

Raise $- 3$ to the power of $2$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Step 15.1.2

Multiply $- 4 \cdot 2 \cdot - 2$ .

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Step 15.1.2.1

Multiply $- 4$ by $2$ .

$u = \frac{3 \pm \sqrt{9 - 8 \cdot - 2}}{2 \cdot 2}$

Step 15.1.2.2

Multiply $- 8$ by $- 2$ .

$u = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 2}$

Step 15.1.3

Add $9$ and $16$ .

$u = \frac{3 \pm \sqrt{25}}{2 \cdot 2}$

Step 15.1.4

Rewrite $25$ as $5^{2}$ .

$u = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 2}$

Step 15.1.5

Pull terms out from under the radical, assuming positive real numbers.

$u = \frac{3 \pm 5}{2 \cdot 2}$

Step 15.2

Multiply $2$ by $2$ .

$u = \frac{3 \pm 5}{4}$

Step 16

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 16.1

Simplify the numerator.

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Step 16.1.1

Raise $- 3$ to the power of $2$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Step 16.1.2

Multiply $- 4 \cdot 2 \cdot - 2$ .

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Step 16.1.2.1

Multiply $- 4$ by $2$ .

$u = \frac{3 \pm \sqrt{9 - 8 \cdot - 2}}{2 \cdot 2}$

Step 16.1.2.2

Multiply $- 8$ by $- 2$ .

$u = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 2}$

Step 16.1.3

Add $9$ and $16$ .

$u = \frac{3 \pm \sqrt{25}}{2 \cdot 2}$

Step 16.1.4

Rewrite $25$ as $5^{2}$ .

$u = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 2}$

Step 16.1.5

Pull terms out from under the radical, assuming positive real numbers.

$u = \frac{3 \pm 5}{2 \cdot 2}$

Step 16.2

Multiply $2$ by $2$ .

$u = \frac{3 \pm 5}{4}$

Step 16.3

Change the $\pm$ to $+$ .

$u = \frac{3 + 5}{4}$

Step 16.4

Add $3$ and $5$ .

$u = \frac{8}{4}$

Step 16.5

Divide $8$ by $4$ .

$u = 2$

Step 17

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 17.1

Simplify the numerator.

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Step 17.1.1

Raise $- 3$ to the power of $2$ .

$u = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 2}}{2 \cdot 2}$

Step 17.1.2

Multiply $- 4 \cdot 2 \cdot - 2$ .

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Step 17.1.2.1

Multiply $- 4$ by $2$ .

$u = \frac{3 \pm \sqrt{9 - 8 \cdot - 2}}{2 \cdot 2}$

Step 17.1.2.2

Multiply $- 8$ by $- 2$ .

$u = \frac{3 \pm \sqrt{9 + 16}}{2 \cdot 2}$

Step 17.1.3

Add $9$ and $16$ .

$u = \frac{3 \pm \sqrt{25}}{2 \cdot 2}$

Step 17.1.4

Rewrite $25$ as $5^{2}$ .

$u = \frac{3 \pm \sqrt{5^{2}}}{2 \cdot 2}$

Step 17.1.5

Pull terms out from under the radical, assuming positive real numbers.

$u = \frac{3 \pm 5}{2 \cdot 2}$

Step 17.2

Multiply $2$ by $2$ .

$u = \frac{3 \pm 5}{4}$

Step 17.3

Change the $\pm$ to $-$ .

$u = \frac{3 - 5}{4}$

Step 17.4

Subtract $5$ from $3$ .

$u = \frac{- 2}{4}$

Step 17.5

Cancel the common factor of $- 2$ and $4$ .

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Step 17.5.1

Factor $2$ out of $- 2$ .

$u = \frac{2 (- 1)}{4}$

Step 17.5.2

Cancel the common factors.

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Step 17.5.2.1

Factor $2$ out of $4$ .

$u = \frac{2 \cdot - 1}{2 \cdot 2}$

Step 17.5.2.2

Cancel the common factor.

$u = \frac{2 \cdot - 1}{2 \cdot 2}$

Step 17.5.2.3

Rewrite the expression.

$u = \frac{- 1}{2}$

Step 17.6

Move the negative in front of the fraction.

$u = - \frac{1}{2}$

Step 18

The final answer is the combination of both solutions.

$u = 2, - \frac{1}{2}$

Step 19

Substitute $\sec (θ)$ for $u$ .

$\sec (θ) = 2, - \frac{1}{2}$

Step 20

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

$\sec (θ) = 2$

$\sec (θ) = - \frac{1}{2}$

Step 21

Solve for $θ$ in $\sec (θ) = 2$ .

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Step 21.1

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

$θ = arcsec (2)$

Step 21.2

Simplify the right side.

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Step 21.2.1

The exact value of $arcsec (2)$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 21.3

The secant function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{3}$

Step 21.4

Simplify $2 π - \frac{π}{3}$ .

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Step 21.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 21.4.2

Combine fractions.

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Step 21.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 21.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Step 21.4.3

Simplify the numerator.

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Step 21.4.3.1

Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Step 21.4.3.2

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Step 21.5

Find the period of $\sec (θ)$ .

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Step 21.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 21.5.2

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

$\frac{2 π}{| 1 |}$

Step 21.5.3

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

$\frac{2 π}{1}$

Step 21.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 21.6

The period of the $\sec (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 22

Solve for $θ$ in $\sec (θ) = - \frac{1}{2}$ .

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Step 22.1

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $- \frac{1}{2}$ does not fall in this range, there is no solution.

No solution

Step 23

List all of the solutions.

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$