Solve for x tan(2x)+tan(x)=0

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

Step 1

Simplify each term.

Apply the tangent double-angle identity.

$\frac{2 \tan (x)}{1 - \tan^{2} (x)} + \tan (x) = 0$

Simplify the denominator.

Rewrite $1$ as $1^{2}$ .

$\frac{2 \tan (x)}{1^{2} - \tan^{2} (x)} + \tan (x) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = \tan (x)$ .

$\frac{2 \tan (x)}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x) = 0$

Step 2

Factor $\tan (x)$ out of $\frac{2 \tan (x)}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x)$ .

Factor $\tan (x)$ out of $\frac{2 \tan (x)}{(1 + \tan (x)) (1 - \tan (x))}$ .

$\tan (x) \frac{2}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x) = 0$

Raise $\tan (x)$ to the power of $1$ .

$\tan (x) \frac{2}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x) = 0$

Factor $\tan (x)$ out of $\tan^{1} (x)$ .

$\tan (x) \frac{2}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x) \cdot 1 = 0$

Factor $\tan (x)$ out of $\tan (x) \frac{2}{(1 + \tan (x)) (1 - \tan (x))} + \tan (x) \cdot 1$ .

$\tan (x) (\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1) = 0$

Step 3

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

$\tan (x) = 0$

$\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1 = 0$

Step 4

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

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

$\tan (x) = 0$

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

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

$x = \arctan (0)$

Simplify the right side.

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

$x = 0$

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.

$x = π + 0$

Add $π$ and $0$ .

$x = π$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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 $\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1$ equal to $0$ and solve for $x$ .

Set $\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1$ equal to $0$ .

$\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1 = 0$

Solve $\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1 = 0$ for $x$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(1 + \tan (x)) (1 - \tan (x)), 1, 1$

The LCM of one and any expression is the expression.

$(1 + \tan (x)) (1 - \tan (x))$

Multiply each term in $\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1 = 0$ by $(1 + \tan (x)) (1 - \tan (x))$ to eliminate the fractions.

Multiply each term in $\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1 = 0$ by $(1 + \tan (x)) (1 - \tan (x))$ .

$\frac{2}{(1 + \tan (x)) (1 - \tan (x))} ((1 + \tan (x)) (1 - \tan (x))) + 1 ((1 + \tan (x)) (1 - \tan (x))) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Simplify the left side.

Simplify each term.

Cancel the common factor of $(1 + \tan (x)) (1 - \tan (x))$ .

Cancel the common factor.

$\frac{2}{(1 + \tan (x)) (1 - \tan (x))} ((1 + \tan (x)) (1 - \tan (x))) + 1 ((1 + \tan (x)) (1 - \tan (x))) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Rewrite the expression.

$2 + 1 ((1 + \tan (x)) (1 - \tan (x))) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Multiply $(1 + \tan (x)) (1 - \tan (x))$ by $1$ .

$2 + (1 + \tan (x)) (1 - \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Expand $(1 + \tan (x)) (1 - \tan (x))$ using the FOIL Method.

Apply the distributive property.

$2 + 1 (1 - \tan (x)) + \tan (x) (1 - \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Apply the distributive property.

$2 + 1 \cdot 1 + 1 (- \tan (x)) + \tan (x) (1 - \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Apply the distributive property.

$2 + 1 \cdot 1 + 1 (- \tan (x)) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$2 + 1 + 1 (- \tan (x)) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Multiply $- \tan (x)$ by $1$ .

$2 + 1 - \tan (x) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Multiply $\tan (x)$ by $1$ .

$2 + 1 - \tan (x) + \tan (x) + \tan (x) (- \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Rewrite using the commutative property of multiplication.

$2 + 1 - \tan (x) + \tan (x) - \tan (x) \tan (x) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Multiply $\tan (x)$ by $\tan (x)$ by adding the exponents.

Move $\tan (x)$ .

$2 + 1 - \tan (x) + \tan (x) - (\tan (x) \tan (x)) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Multiply $\tan (x)$ by $\tan (x)$ .

$2 + 1 - \tan (x) + \tan (x) - \tan^{2} (x) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Add $- \tan (x)$ and $\tan (x)$ .

$2 + 1 + 0 - \tan^{2} (x) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Add $1$ and $0$ .

$2 + 1 - \tan^{2} (x) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Add $2$ and $1$ .

$3 - \tan^{2} (x) = 0 ((1 + \tan (x)) (1 - \tan (x)))$

Simplify the right side.

Expand $(1 + \tan (x)) (1 - \tan (x))$ using the FOIL Method.

Apply the distributive property.

$3 - \tan^{2} (x) = 0 (1 (1 - \tan (x)) + \tan (x) (1 - \tan (x)))$

Apply the distributive property.

$3 - \tan^{2} (x) = 0 (1 \cdot 1 + 1 (- \tan (x)) + \tan (x) (1 - \tan (x)))$

Apply the distributive property.

$3 - \tan^{2} (x) = 0 (1 \cdot 1 + 1 (- \tan (x)) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)))$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$3 - \tan^{2} (x) = 0 (1 + 1 (- \tan (x)) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)))$

Multiply $- \tan (x)$ by $1$ .

$3 - \tan^{2} (x) = 0 (1 - \tan (x) + \tan (x) \cdot 1 + \tan (x) (- \tan (x)))$

Multiply $\tan (x)$ by $1$ .

$3 - \tan^{2} (x) = 0 (1 - \tan (x) + \tan (x) + \tan (x) (- \tan (x)))$

Rewrite using the commutative property of multiplication.

$3 - \tan^{2} (x) = 0 (1 - \tan (x) + \tan (x) - \tan (x) \tan (x))$

Multiply $\tan (x)$ by $\tan (x)$ by adding the exponents.

Move $\tan (x)$ .

$3 - \tan^{2} (x) = 0 (1 - \tan (x) + \tan (x) - (\tan (x) \tan (x)))$

Multiply $\tan (x)$ by $\tan (x)$ .

$3 - \tan^{2} (x) = 0 (1 - \tan (x) + \tan (x) - \tan^{2} (x))$

Add $- \tan (x)$ and $\tan (x)$ .

$3 - \tan^{2} (x) = 0 (1 + 0 - \tan^{2} (x))$

Add $1$ and $0$ .

$3 - \tan^{2} (x) = 0 (1 - \tan^{2} (x))$

Multiply $0$ by $1 - \tan^{2} (x)$ .

$3 - \tan^{2} (x) = 0$

Solve the equation.

Subtract $3$ from both sides of the equation.

$- \tan^{2} (x) = - 3$

Divide each term in $- \tan^{2} (x) = - 3$ by $- 1$ and simplify.

Divide each term in $- \tan^{2} (x) = - 3$ by $- 1$ .

$\frac{- \tan^{2} (x)}{- 1} = \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\tan^{2} (x)}{1} = \frac{- 3}{- 1}$

Divide $\tan^{2} (x)$ by $1$ .

$\tan^{2} (x) = \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$\tan^{2} (x) = 3$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

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

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

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

Next, use the negative value of the $\pm$ to find the second solution.

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

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

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

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

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

Simplify the right side.

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

$x = \frac{π}{3}$

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.

$x = π + \frac{π}{3}$

Simplify $π + \frac{π}{3}$ .

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

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

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$x = \frac{π \cdot 3 + π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$x = \frac{3 \cdot π + π}{3}$

Add $3 π$ and $π$ .

$x = \frac{4 π}{3}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

Solve for $x$ in $\tan (x) = - \sqrt{3}$ .

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

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

Simplify the right side.

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

$x = - \frac{π}{3}$

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.

$x = - \frac{π}{3} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{3} - π$ .

$x = - \frac{π}{3} - π + 2 π$

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

$x = \frac{2 π}{3}$

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

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

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

$- \frac{π}{3} + π$

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

$π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

$\frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{π \cdot 3 - π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$\frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 π$ .

$\frac{2 π}{3}$

List the new angles.

$x = \frac{2 π}{3}$

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 $n$

List all of the solutions.

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

Consolidate the solutions.

Consolidate $\frac{π}{3} + π n$ and $\frac{4 π}{3} + π n$ to $\frac{π}{3} + π n$ .

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

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

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

Step 6

The final solution is all the values that make $\tan (x) (\frac{2}{(1 + \tan (x)) (1 - \tan (x))} + 1) = 0$ true.

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

Step 7

Consolidate the answers.

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

Solve for x tan(2x)+tan(x)=0

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