Trigonometry Examples

$\cos (θ) - \tan (θ) \cos (θ) = 0$

Step 1

Factor $\cos (θ)$ out of $\cos (θ) - \tan (θ) \cos (θ)$ .

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Step 1.1

Raise $\cos (θ)$ to the power of $1$ .

$\cos (θ) - \tan (θ) \cos (θ) = 0$

Step 1.2

Factor $\cos (θ)$ out of $\cos^{1} (θ)$ .

$\cos (θ) \cdot 1 - \tan (θ) \cos (θ) = 0$

Step 1.3

Factor $\cos (θ)$ out of $- \tan (θ) \cos (θ)$ .

$\cos (θ) \cdot 1 + \cos (θ) (- \tan (θ)) = 0$

Step 1.4

Factor $\cos (θ)$ out of $\cos (θ) \cdot 1 + \cos (θ) (- \tan (θ))$ .

$\cos (θ) (1 - \tan (θ)) = 0$

Step 2

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

$\cos (θ) = 0$

$1 - \tan (θ) = 0$

Step 3

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

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Step 3.1

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Step 3.2

Solve $\cos (θ) = 0$ for $θ$ .

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Step 3.2.1

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

$θ = \arccos (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

Step 3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{2}$

Step 3.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 3.2.4.1

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

$θ = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.2.4.2

Combine fractions.

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Step 3.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$θ = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.2.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Step 3.2.4.3

Simplify the numerator.

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Step 3.2.4.3.1

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Step 3.2.4.3.2

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Step 3.2.5

Find the period of $\cos (θ)$ .

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Step 3.2.5.1

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

$\frac{2 π}{| b |}$

Step 3.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.5.3

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

$\frac{2 π}{1}$

Step 3.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.6

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

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

Step 4

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

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Step 4.1

Set $1 - \tan (θ)$ equal to $0$ .

$1 - \tan (θ) = 0$

Step 4.2

Solve $1 - \tan (θ) = 0$ for $θ$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$- \tan (θ) = - 1$

Step 4.2.2

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

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Step 4.2.2.1

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

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 4.2.2.2.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{- 1}{- 1}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Divide $- 1$ by $- 1$ .

$\tan (θ) = 1$

Step 4.2.3

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

$θ = \arctan (1)$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$θ = \frac{π}{4}$

Step 4.2.5

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.

$θ = π + \frac{π}{4}$

Step 4.2.6

Simplify $π + \frac{π}{4}$ .

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Step 4.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$θ = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 4.2.6.2

Combine fractions.

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Step 4.2.6.2.1

Combine $π$ and $\frac{4}{4}$ .

$θ = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 4.2.6.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 + π}{4}$

Step 4.2.6.3

Simplify the numerator.

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Step 4.2.6.3.1

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π + π}{4}$

Step 4.2.6.3.2

Add $4 π$ and $π$ .

$θ = \frac{5 π}{4}$

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 $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 4.2.7.2

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

$\frac{π}{| 1 |}$

Step 4.2.7.3

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

$\frac{π}{1}$

Step 4.2.7.4

Divide $π$ by $1$ .

$π$

Step 4.2.8

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

$θ = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 5

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

$θ = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$