Graph sin(theta)>0 , sec(theta)>0

$\sin (θ) > 0$ , $\sec (θ) > 0$

Solve for $θ$ .

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

$θ > \arcsin (0)$

Simplify the right side.

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

$θ > 0$

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

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

Find the period of $\sin (θ)$ .

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 $\sin (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Use each root to create test intervals.

$0 < θ < π$

$π < θ < 2 π$

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 $0 < θ < π$ to see if it makes the inequality true.

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

$θ = 2$

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

$\sin (2) > 0$

The left side $0.90929742$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

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

$θ = 5$

Replace $θ$ with $5$ in the original inequality.

$\sin (5) > 0$

The left side $- 0.95892427$ is not greater than the right side $0$ , which means that the given statement is false.

False

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

$0 < θ < π$ True

$π < θ < 2 π$ False

$0 < θ < π$ True

$π < θ < 2 π$ False

The solution consists of all of the true intervals.

$0 + 2 π n < θ < π + 2 π n$ , for any integer $n$

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Plot each graph on the same coordinate system.

$\sin (θ) > 0$

$\sec (θ) > 0$

Graph sin(theta)>0 , sec(theta)>0

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