Solve for x 2tan(x)=1-tan(x)^2

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

Substitute $u$ for $\tan (x)$ .

$2 (u) = 1 - {(u)}^{2}$

Add $- u^{2}$ to both sides of the equation.

$2 u + u^{2} = 1$

Move $1$ to the left side of the equation by subtracting it from both sides.

$2 u + u^{2} - 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $u$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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

$u = \frac{- 2 \pm \sqrt{4 - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Multiply $- 1$ by $1$ .

$u = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Multiply $- 4$ by $- 1$ .

$u = \frac{- 2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Add $4$ and $4$ .

$u = \frac{- 2 \pm \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

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Factor $4$ out of $8$ .

$u = \frac{- 2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

$u = \frac{- 2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$u = \frac{- 2 \pm 2 \sqrt{2}}{2}$

Simplify $\frac{- 2 \pm 2 \sqrt{2}}{2}$ .

$u = - 1 \pm \sqrt{2}$

The final answer is the combination of both solutions.

$u = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

Substitute $\tan (x)$ for $u$ .

$\tan (x) = - 1 + \sqrt{2}, - 1 - \sqrt{2}$

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

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

$\tan (x) = - 1 - \sqrt{2}$

Set up the equation to solve for $x$ .

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

Solve the equation for $x$ .

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Convert the right side of the equation to its decimal equivalent.

$\tan (x) = 0.41421356$

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

$x = \arctan (0.41421356)$

Evaluate $\arctan (0.41421356)$ .

$x = \frac{π}{8}$

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.

$x = π + \frac{π}{8}$

Simplify $π + \frac{π}{8}$ .

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To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x = \frac{π}{1} \cdot \frac{8}{8} + \frac{π}{8}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = \frac{π \cdot 8}{1 \cdot 8} + \frac{π}{8}$

Multiply $8$ by $1$ .

$x = \frac{π \cdot 8}{8} + \frac{π}{8}$

Combine the numerators over the common denominator.

$x = \frac{π \cdot 8 + π}{8}$

Simplify the numerator.

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Move $8$ to the left of $π$ .

$x = \frac{8 \cdot π + π}{8}$

Add $8 π$ and $π$ .

$x = \frac{9 π}{8}$

Find the period.

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

Solve the equation.

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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.

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

Set up the equation to solve for $x$ .

$\tan (x) = - 1 - \sqrt{2}$

Solve the equation for $x$ .

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Convert the right side of the equation to its decimal equivalent.

$\tan (x) = - 2.41421356$

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

$x = \arctan (- 2.41421356)$

Evaluate $\arctan (- 2.41421356)$ .

$x = - \frac{3 π}{8}$

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.

$x = - \frac{3 π}{8} - π$

Simplify the expression to find the second solution.

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Simplify $- \frac{3 π}{8} - π$ .

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To write $- \frac{- π}{1}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x = - \frac{3 π}{8} + \frac{- π}{1} \cdot \frac{8}{8}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$x = - \frac{3 π}{8} + \frac{- π \cdot 8}{1 \cdot 8}$

Multiply $8$ by $1$ .

$x = - \frac{3 π}{8} + \frac{- π \cdot 8}{8}$

Combine the numerators over the common denominator.

$x = \frac{- 3 π - π \cdot 8}{8}$

Simplify the numerator.

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Multiply $8$ by $- 1$ .

$x = \frac{- 3 π - 8 π}{8}$

Subtract $8 π$ from $- 3 π$ .

$x = \frac{- 11 π}{8}$

Move the negative in front of the fraction.

$x = - \frac{11 π}{8}$

Add $2 π$ to $- \frac{11 π}{8}$ .

$x = - \frac{11 π}{8} + 2 π$

The resulting angle of $\frac{5 π}{8}$ is positive and coterminal with $- \frac{11 π}{8}$ .

$x = \frac{5 π}{8}$

Find the period.

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

Solve the equation.

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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.

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Add $π$ to $- \frac{3 π}{8}$ to find the positive angle.

$- \frac{3 π}{8} + π$

To write $\frac{π}{1}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$\frac{π}{1} \cdot \frac{8}{8} - \frac{3 π}{8}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{π \cdot 8}{1 \cdot 8} - \frac{3 π}{8}$

Multiply $8$ by $1$ .

$\frac{π \cdot 8}{8} - \frac{3 π}{8}$

Combine the numerators over the common denominator.

$\frac{π \cdot 8 - 3 π}{8}$

Simplify the numerator.

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Move $8$ to the left of $π$ .

$\frac{8 \cdot π - 3 π}{8}$

Subtract $3 π$ from $8 π$ .

$\frac{5 π}{8}$

List the new angles.

$x = \frac{5 π}{8}$

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

$x = \frac{5 π}{8} + π n, \frac{5 π}{8} + π n$ , for any integer $n$

List all of the results found in the previous steps.

$x = \frac{π}{8} + π n, \frac{5 π}{8} + π n, \frac{9 π}{8} + π n$ , for any integer $n$

The complete solution is the set of all solutions.

$x = \frac{π}{8} + π n, \frac{9 π}{8} + π n, \frac{5 π}{8} + π n, \frac{5 π}{8} + π n$ , for any integer $n$

Consolidate the answers.

$x = \frac{π}{8} + \frac{π n}{2}$ , for any integer $n$

Solve for x 2tan(x)=1-tan(x)^2

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