Trigonometry Examples

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

Step 1

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

$\tan (x - \frac{1}{3}) = \pm \sqrt{0}$

Step 2

Simplify $\pm \sqrt{0}$ .

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Step 2.1

Rewrite $0$ as $0^{2}$ .

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

Step 2.2

Pull terms out from under the radical, assuming positive real numbers.

$\tan (x - \frac{1}{3}) = \pm 0$

Step 2.3

Plus or minus $0$ is $0$ .

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

Step 3

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

$x - \frac{1}{3} = \arctan (0)$

Step 4

Simplify the right side.

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Step 4.1

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

$x - \frac{1}{3} = 0$

Step 5

Add $\frac{1}{3}$ to both sides of the equation.

$x = \frac{1}{3}$

Step 6

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.

$x - \frac{1}{3} = π + 0$

Step 7

Solve for $x$ .

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Step 7.1

Add $π$ and $0$ .

$x - \frac{1}{3} = π$

Step 7.2

Add $\frac{1}{3}$ to both sides of the equation.

$x = π + \frac{1}{3}$

Step 8

Find the period of $\tan (x - \frac{1}{3})$ .

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Step 8.1

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

$\frac{π}{| b |}$

Step 8.2

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

$\frac{π}{| 1 |}$

Step 8.3

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

$\frac{π}{1}$

Step 8.4

Divide $π$ by $1$ .

$π$

Step 9

The period of the $\tan (x - \frac{1}{3})$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \frac{1}{3} + π n, π + \frac{1}{3} + π n$ , for any integer $n$

Step 10

Consolidate $\frac{1}{3} + π n$ and $π + \frac{1}{3} + π n$ to $\frac{1}{3} + π n$ .

$x = \frac{1}{3} + π n$ , for any integer $n$