Graph tan(theta)<0 , cos(theta)<0

$\tan (θ) < 0$ , $\cos (θ) < 0$

Solve for $θ$ .

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

$θ < \arctan (0)$

Simplify the right side.

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

$θ < 0$

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.

$θ = π + 0$

Add $π$ and $0$ .

$θ = π$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

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

Consolidate the answers.

$θ = π n$ , for any integer $n$

Find the domain of $\tan (θ)$ .

Set the argument in $\tan (θ)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

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

The domain is all values of $θ$ that make the expression defined.

${θ | θ \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Use each root to create test intervals.

$0 < θ < \frac{π}{2}$

$\frac{π}{2} < θ < π$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $0 < θ < \frac{π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < θ < \frac{π}{2}$ and see if this value makes the original inequality true.

$θ = 1$

Replace $θ$ with $1$ in the original inequality.

$\tan (1) < 0$

The left side $1.55740772$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $\frac{π}{2} < θ < π$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < θ < π$ and see if this value makes the original inequality true.

$θ = 2$

Replace $θ$ with $2$ in the original inequality.

$\tan (2) < 0$

The left side $- 2.18503986$ is less than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$0 < θ < \frac{π}{2}$ False

$\frac{π}{2} < θ < π$ True

$0 < θ < \frac{π}{2}$ False

$\frac{π}{2} < θ < π$ True

The solution consists of all of the true intervals.

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

Solve for $θ$ .

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

$θ < \arccos (0)$

Simplify the right side.

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

$θ < \frac{π}{2}$

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.

$θ = 2 π - \frac{π}{2}$

Simplify $2 π - \frac{π}{2}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Find the period of $\cos (θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

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 $n$

Consolidate the answers.

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

The domain is all values of $θ$ that make the expression defined.

${θ | θ \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Use each root to create test intervals.

$\frac{π}{2} < θ < \frac{3 π}{2}$

$\frac{3 π}{2} < θ < \frac{5 π}{2}$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $\frac{π}{2} < θ < \frac{3 π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < θ < \frac{3 π}{2}$ and see if this value makes the original inequality true.

$θ = 3$

Replace $θ$ with $3$ in the original inequality.

$\tan (3) < 0$

The left side $- 0.14254654$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{3 π}{2} < θ < \frac{5 π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{3 π}{2} < θ < \frac{5 π}{2}$ and see if this value makes the original inequality true.

$θ = 6$

Replace $θ$ with $6$ in the original inequality.

$\tan (6) < 0$

The left side $- 0.29100619$ is less than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$\frac{π}{2} < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < \frac{5 π}{2}$ True

$\frac{π}{2} < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < \frac{5 π}{2}$ True

The solution consists of all of the true intervals.

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

Plot each graph on the same coordinate system.

$\tan (θ) < 0$

$\cos (θ) < 0$

Graph tan(theta)<0 , cos(theta)<0

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