Precalculus Examples

$(\tan (x) + \sqrt{3}) (\sec (x) - 2) = 0$

Step 1

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

$\tan (x) + \sqrt{3} = 0$

$\sec (x) - 2 = 0$

Step 2

Set $\tan (x) + \sqrt{3}$ equal to $0$ and solve for $x$ .

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Step 2.1

Set $\tan (x) + \sqrt{3}$ equal to $0$ .

$\tan (x) + \sqrt{3} = 0$

Step 2.2

Solve $\tan (x) + \sqrt{3} = 0$ for $x$ .

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Step 2.2.1

Subtract $\sqrt{3}$ from both sides of the equation.

$\tan (x) = - \sqrt{3}$

Step 2.2.2

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

$x = \arctan (- \sqrt{3})$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

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

$x = - \frac{π}{3}$

Step 2.2.4

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

$x = - \frac{π}{3} - π$

Step 2.2.5

Simplify the expression to find the second solution.

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Step 2.2.5.1

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

Step 2.2.5.2

The resulting angle of $\frac{2 π}{3}$ is positive and coterminal with $- \frac{π}{3} - π$ .

$x = \frac{2 π}{3}$

Step 2.2.6

Find the period of $\tan (x)$ .

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Step 2.2.6.1

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

$\frac{π}{| b |}$

Step 2.2.6.2

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

$\frac{π}{| 1 |}$

Step 2.2.6.3

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

$\frac{π}{1}$

Step 2.2.6.4

Divide $π$ by $1$ .

$π$

Step 2.2.7

Add $π$ to every negative angle to get positive angles.

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Step 2.2.7.1

Add $π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + π$

Step 2.2.7.2

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

$π \cdot \frac{3}{3} - \frac{π}{3}$

Step 2.2.7.3

Combine fractions.

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Step 2.2.7.3.1

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Step 2.2.7.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Step 2.2.7.4

Simplify the numerator.

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Step 2.2.7.4.1

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Step 2.2.7.4.2

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

Step 2.2.7.5

List the new angles.

$x = \frac{2 π}{3}$

Step 2.2.8

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

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

Step 3

Set $\sec (x) - 2$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\sec (x) - 2$ equal to $0$ .

$\sec (x) - 2 = 0$

Step 3.2

Solve $\sec (x) - 2 = 0$ for $x$ .

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Step 3.2.1

Add $2$ to both sides of the equation.

$\sec (x) = 2$

Step 3.2.2

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

$x = arcsec (2)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

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

$x = \frac{π}{3}$

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 $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{3}$

Step 3.2.5

Simplify $2 π - \frac{π}{3}$ .

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Step 3.2.5.1

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

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

Step 3.2.5.2

Combine fractions.

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Step 3.2.5.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 3.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 3.2.5.3

Simplify the numerator.

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Step 3.2.5.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 3.2.5.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 3.2.6

Find the period of $\sec (x)$ .

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

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

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

Step 4

The final solution is all the values that make $(\tan (x) + \sqrt{3}) (\sec (x) - 2) = 0$ true.

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

Step 5

Consolidate $\frac{2 π}{3} + π n$ and $\frac{5 π}{3} + 2 π n$ to $\frac{2 π}{3} + π n$ .

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