Trigonometry Examples

$2 \tan^{2} (θ) + \tan (θ) = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = \tan (θ)$ . Substitute $u$ for all occurrences of $\tan (θ)$ .

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

Step 1.2

Factor $u$ out of $2 u^{2} + u$ .

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Step 1.2.1

Factor $u$ out of $2 u^{2}$ .

$u (2 u) + u = 0$

Step 1.2.2

Raise $u$ to the power of $1$ .

$u (2 u) + u = 0$

Step 1.2.3

Factor $u$ out of $u^{1}$ .

$u (2 u) + u \cdot 1 = 0$

Step 1.2.4

Factor $u$ out of $u (2 u) + u \cdot 1$ .

$u (2 u + 1) = 0$

Step 1.3

Replace all occurrences of $u$ with $\tan (θ)$ .

$\tan (θ) (2 \tan (θ) + 1) = 0$

Step 2

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

$\tan (θ) = 0$

$2 \tan (θ) + 1 = 0$

Step 3

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

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Step 3.1

Set $\tan (θ)$ equal to $0$ .

$\tan (θ) = 0$

Step 3.2

Solve $\tan (θ) = 0$ for $θ$ .

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Step 3.2.1

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

$θ = \arctan (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$θ = 0$

Step 3.2.3

The tangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 0$

Step 3.2.4

Add $180$ and $0$ .

$θ = 180$

Step 3.2.5

Find the period of $\tan (θ)$ .

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Step 3.2.5.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 3.2.5.2

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

$\frac{180}{| 1 |}$

Step 3.2.5.3

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

$\frac{180}{1}$

Step 3.2.5.4

Divide $180$ by $1$ .

$180$

Step 3.2.6

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 180 n, 180 + 180 n$ , for any integer $n$

Step 4

Set $2 \tan (θ) + 1$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $2 \tan (θ) + 1$ equal to $0$ .

$2 \tan (θ) + 1 = 0$

Step 4.2

Solve $2 \tan (θ) + 1 = 0$ for $θ$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$2 \tan (θ) = - 1$

Step 4.2.2

Divide each term in $2 \tan (θ) = - 1$ by $2$ and simplify.

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Step 4.2.2.1

Divide each term in $2 \tan (θ) = - 1$ by $2$ .

$\frac{2 \tan (θ)}{2} = \frac{- 1}{2}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{2 \tan (θ)}{2} = \frac{- 1}{2}$

Step 4.2.2.2.1.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{- 1}{2}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Move the negative in front of the fraction.

$\tan (θ) = - \frac{1}{2}$

Step 4.2.3

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

$θ = \arctan (- \frac{1}{2})$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

Evaluate $\arctan (- \frac{1}{2})$ .

$θ = - 26.56505117$

Step 4.2.5

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 26.56505117 - 180$

Step 4.2.6

Simplify the expression to find the second solution.

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Step 4.2.6.1

Add $360 °$ to $- 26.56505117 - 180 °$ .

$θ = - 26.56505117 - 180 ° + 360 °$

Step 4.2.6.2

The resulting angle of $153.43494882 °$ is positive and coterminal with $- 26.56505117 - 180$ .

$θ = 153.43494882 °$

Step 4.2.7

Find the period of $\tan (θ)$ .

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Step 4.2.7.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 4.2.7.2

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

$\frac{180}{| 1 |}$

Step 4.2.7.3

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

$\frac{180}{1}$

Step 4.2.7.4

Divide $180$ by $1$ .

$180$

Step 4.2.8

Add $180$ to every negative angle to get positive angles.

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Step 4.2.8.1

Add $180$ to $- 26.56505117$ to find the positive angle.

$- 26.56505117 + 180$

Step 4.2.8.2

Subtract $26.56505117$ from $180$ .

$153.43494882$

Step 4.2.8.3

List the new angles.

$θ = 153.43494882$

Step 4.2.9

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 153.43494882 + 180 n, 153.43494882 + 180 n$ , for any integer $n$

Step 5

The final solution is all the values that make $\tan (θ) (2 \tan (θ) + 1) = 0$ true.

$θ = 180 n, 180 + 180 n, 153.43494882 + 180 n$ , for any integer $n$

Step 6

Consolidate $180 n$ and $180 + 180 n$ to $180 n$ .

$θ = 180 n, 153.43494882 + 180 n$ , for any integer $n$