Find All Complex Solutions tan(theta)^2=-3/2sec(theta)

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

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

Rewrite.

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

Simplify by adding zeros.

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the expression.

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

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

Move $3$ to the left of $u$ .

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

Step 11

Add $\frac{3 u}{2}$ to 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.

Apply the distributive property.

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

Simplify.

Multiply $2$ by $- 1$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 2$ .

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

Add $9$ and $16$ .

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

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

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

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

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

Multiply $2$ by $2$ .

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

Step 16

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 2$ .

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

Add $9$ and $16$ .

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

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

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

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

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

Multiply $2$ by $2$ .

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

Change the $\pm$ to $+$ .

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

Add $- 3$ and $5$ .

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

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

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 17

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $2$ .

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

Multiply $- 8$ by $- 2$ .

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

Add $9$ and $16$ .

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

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

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

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

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

Multiply $2$ by $2$ .

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

Change the $\pm$ to $-$ .

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

Subtract $5$ from $- 3$ .

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

Divide $- 8$ by $4$ .

$u = - 2$

Step 18

The final answer is the combination of both solutions.

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

Step 19

Substitute $\sec (θ)$ for $u$ .

$\sec (θ) = \frac{1}{2}, - 2$

Step 20

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

$\sec (θ) = \frac{1}{2}$

$\sec (θ) = - 2$

Step 21

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

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 22

Solve for $θ$ in $\sec (θ) = - 2$ .

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

$θ = arcsec (- 2)$

Simplify the right side.

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

$θ = \frac{2 π}{3}$

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

$θ = 2 π - \frac{2 π}{3}$

Simplify $2 π - \frac{2 π}{3}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

Find the period of $\sec (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Step 23

List all of the solutions.

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Find All Complex Solutions tan(theta)^2=-3/2sec(theta)

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