Trigonometry Examples

$\tan (x) \sec (x) = - 2 \tan (x)$

Step 1

Add $2 \tan (x)$ to both sides of the equation.

$\tan (x) \sec (x) + 2 \tan (x) = 0$

Step 2

Factor $\tan (x)$ out of $\tan (x) \sec (x) + 2 \tan (x)$ .

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Step 2.1

Factor $\tan (x)$ out of $2 \tan (x)$ .

$\tan (x) \sec (x) + \tan (x) \cdot 2 = 0$

Step 2.2

Factor $\tan (x)$ out of $\tan (x) \sec (x) + \tan (x) \cdot 2$ .

$\tan (x) (\sec (x) + 2) = 0$

Step 3

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

$\tan (x) = 0$

$\sec (x) + 2 = 0$

Step 4

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

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Step 4.1

Set $\tan (x)$ equal to $0$ .

$\tan (x) = 0$

Step 4.2

Solve $\tan (x) = 0$ for $x$ .

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Step 4.2.1

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

$x = \arctan (0)$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

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

$x = 0$

Step 4.2.3

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

$x = π + 0$

Step 4.2.4

Add $π$ and $0$ .

$x = π$

Step 4.2.5

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

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Step 4.2.5.1

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

$\frac{π}{| b |}$

Step 4.2.5.2

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

$\frac{π}{| 1 |}$

Step 4.2.5.3

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

$\frac{π}{1}$

Step 4.2.5.4

Divide $π$ by $1$ .

$π$

Step 4.2.6

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

$x = π n, π + π n$ , for any integer $n$

Step 5

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

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Step 5.1

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

$\sec (x) + 2 = 0$

Step 5.2

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

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Step 5.2.1

Subtract $2$ from both sides of the equation.

$\sec (x) = - 2$

Step 5.2.2

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

$x = arcsec (- 2)$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

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

$x = \frac{2 π}{3}$

Step 5.2.4

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.

$x = 2 π - \frac{2 π}{3}$

Step 5.2.5

Simplify $2 π - \frac{2 π}{3}$ .

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Step 5.2.5.1

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

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

Step 5.2.5.2

Combine fractions.

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Step 5.2.5.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 5.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 5.2.5.3

Simplify the numerator.

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Step 5.2.5.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 5.2.5.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 5.2.6

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

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Step 5.2.6.1

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

$\frac{2 π}{| b |}$

Step 5.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 5.2.6.3

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

$\frac{2 π}{1}$

Step 5.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 5.2.7

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

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

Step 6

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

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

Step 7

Consolidate $π n$ and $π + π n$ to $π n$ .

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