Trigonometry Examples

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

Step 1

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

$\tan (x + 1) = 0$

$\cos (x - 1) = 0$

Step 2

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

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Step 2.1

Set $\tan (x + 1)$ equal to $0$ .

$\tan (x + 1) = 0$

Step 2.2

Solve $\tan (x + 1) = 0$ for $x$ .

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Step 2.2.1

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

$x + 1 = \arctan (0)$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

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

$x + 1 = 0$

Step 2.2.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.2.4

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.

$x + 1 = π + 0$

Step 2.2.5

Solve for $x$ .

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Step 2.2.5.1

Add $π$ and $0$ .

$x + 1 = π$

Step 2.2.5.2

Subtract $1$ from both sides of the equation.

$x = π - 1$

Step 2.2.6

Find the period of $\tan (x + 1)$ .

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Step 2.2.6.1

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

$\frac{π}{| b |}$

Step 2.2.6.2

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

$\frac{π}{| 1 |}$

Step 2.2.6.3

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

$\frac{π}{1}$

Step 2.2.6.4

Divide $π$ by $1$ .

$π$

Step 2.2.7

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

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Step 2.2.7.1

Add $π$ to $- 1$ to find the positive angle.

$- 1 + π$

Step 2.2.7.2

List the new angles.

$x = - 1 + π$

Step 2.2.8

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

$x = - 1 + π + π n, π - 1 + π + π n$ , for any integer $n$

Step 3

Set $\cos (x - 1)$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\cos (x - 1)$ equal to $0$ .

$\cos (x - 1) = 0$

Step 3.2

Solve $\cos (x - 1) = 0$ for $x$ .

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Step 3.2.1

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

$x - 1 = \arccos (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$x - 1 = \frac{π}{2}$

Step 3.2.3

Add $1$ to both sides of the equation.

$x = \frac{π}{2} + 1$

Step 3.2.4

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.

$x - 1 = 2 π - \frac{π}{2}$

Step 3.2.5

Solve for $x$ .

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Step 3.2.5.1

Simplify $2 π - \frac{π}{2}$ .

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Step 3.2.5.1.1

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

$x - 1 = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.2.5.1.2

Combine fractions.

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Step 3.2.5.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x - 1 = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.2.5.1.2.2

Combine the numerators over the common denominator.

$x - 1 = \frac{2 π \cdot 2 - π}{2}$

Step 3.2.5.1.3

Simplify the numerator.

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Step 3.2.5.1.3.1

Multiply $2$ by $2$ .

$x - 1 = \frac{4 π - π}{2}$

Step 3.2.5.1.3.2

Subtract $π$ from $4 π$ .

$x - 1 = \frac{3 π}{2}$

Step 3.2.5.2

Add $1$ to both sides of the equation.

$x = \frac{3 π}{2} + 1$

Step 3.2.6

Find the period of $\cos (x - 1)$ .

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

The period of the $\cos (x - 1)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{2} + 1 + 2 π n, \frac{3 π}{2} + 1 + 2 π n$ , for any integer $n$

Step 4

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

$x = - 1 + π + π n, π - 1 + π + π n, \frac{π}{2} + 1 + 2 π n, \frac{3 π}{2} + 1 + 2 π n$ , for any integer $n$

Step 5

Consolidate the answers.

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Step 5.1

Consolidate $- 1 + π + π n$ and $π - 1 + π + π n$ to $- 1 + π + π n$ .

$x = - 1 + π + π n, \frac{π}{2} + 1 + 2 π n, \frac{3 π}{2} + 1 + 2 π n$ , for any integer $n$

Step 5.2

Consolidate $\frac{π}{2} + 1 + 2 π n$ and $\frac{3 π}{2} + 1 + 2 π n$ to $\frac{π}{2} + 1 + π n$ .

$x = - 1 + π + π n, \frac{π}{2} + 1 + π n$ , for any integer $n$