Solve for x sec((3x)/2)=-2

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Trigonometry

$\sec (\frac{3 x}{2}) = - 2$

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

$\frac{3 x}{2} = arcsec (- 2)$

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

$\frac{3 x}{2} = \frac{2 π}{3}$

Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 x}{2} = \frac{2}{3} \cdot \frac{2 π}{3}$

Simplify both sides of the equation.

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Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 x}{2} = \frac{2}{3} \cdot \frac{2 π}{3}$

Rewrite the expression.

$\frac{1}{3} \cdot (3 x) = \frac{2}{3} \cdot \frac{2 π}{3}$

Cancel the common factor of $3$ .

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Factor $3$ out of $3 x$ .

$\frac{1}{3} \cdot (3 (x)) = \frac{2}{3} \cdot \frac{2 π}{3}$

Cancel the common factor.

$\frac{1}{3} \cdot (3 x) = \frac{2}{3} \cdot \frac{2 π}{3}$

Rewrite the expression.

$x = \frac{2}{3} \cdot \frac{2 π}{3}$

Multiply $\frac{2}{3} \cdot \frac{2 π}{3}$ .

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Multiply $\frac{2}{3}$ and $\frac{2 π}{3}$ .

$x = \frac{2 (2 π)}{3 \cdot 3}$

Multiply $2$ by $2$ .

$x = \frac{4 π}{3 \cdot 3}$

Multiply $3$ by $3$ .

$x = \frac{4 π}{9}$

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.

$\frac{3 x}{2} = 2 π - \frac{2 π}{3}$

Simplify the expression to find the second solution.

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Multiply both sides of the equation by $\frac{2}{3}$ .

$\frac{2}{3} \cdot \frac{3 x}{2} = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Simplify both sides of the equation.

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Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2}{3} \cdot \frac{3 x}{2} = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Rewrite the expression.

$\frac{1}{3} \cdot (3 x) = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Cancel the common factor of $3$ .

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Factor $3$ out of $3 x$ .

$\frac{1}{3} \cdot (3 (x)) = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Cancel the common factor.

$\frac{1}{3} \cdot (3 x) = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Rewrite the expression.

$x = \frac{2}{3} \cdot (2 π - \frac{2 π}{3})$

Simplify $\frac{2}{3} \cdot (2 π - \frac{2 π}{3})$ .

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To write $\frac{2 π}{1}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x = \frac{2}{3} \cdot (\frac{2 π}{1} \cdot \frac{3}{3} - \frac{2 π}{3})$

Write each expression with a common denominator of $3$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{2}{3} \cdot (\frac{2 π \cdot 3}{1 \cdot 3} - \frac{2 π}{3})$

Multiply $3$ by $1$ .

$x = \frac{2}{3} \cdot (\frac{2 π \cdot 3}{3} - \frac{2 π}{3})$

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Multiply $3$ by $2$ .

$x = \frac{2}{3} \cdot \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$x = \frac{2}{3} \cdot \frac{4 π}{3}$

Multiply $\frac{2}{3} \cdot \frac{4 π}{3}$ .

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Multiply $\frac{2}{3}$ and $\frac{4 π}{3}$ .

$x = \frac{2 (4 π)}{3 \cdot 3}$

Multiply $4$ by $2$ .

$x = \frac{8 π}{3 \cdot 3}$

Multiply $3$ by $3$ .

$x = \frac{8 π}{9}$

Find the period.

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

$\frac{2 π}{| b |}$

Replace $b$ with $\frac{3}{2}$ in the formula for period.

$\frac{2 π}{| \frac{3}{2} |}$

Solve the equation.

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$\frac{3}{2}$ is approximately $1.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{3}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$2 π \frac{2}{3}$

Multiply $2 π \frac{2}{3}$ .

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

$\frac{2 \cdot 2}{3} π$

Multiply $2$ by $2$ .

$\frac{4}{3} π$

Combine $\frac{4}{3}$ and $π$ .

$\frac{4 π}{3}$

The period of the $\sec (\frac{3 x}{2})$ function is $\frac{4 π}{3}$ so values will repeat every $\frac{4 π}{3}$ radians in both directions.

$x = \frac{4 π}{9} + \frac{4 π n}{3}, \frac{8 π}{9} + \frac{4 π n}{3}$ , for any integer $n$

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