Trigonometry Examples

$\arcsin (x) > \frac{π}{3}$

Step 1

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

$x > \sin (\frac{π}{3})$

Step 2

Simplify the right side.

Tap for more steps...

Step 2.1

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$x > \frac{\sqrt{3}}{2}$

Step 3

Find the domain of $\arcsin (x) - \frac{π}{3}$ .

Tap for more steps...

Step 3.1

Set the argument in $\arcsin (x)$ greater than or equal to $- 1$ to find where the expression is defined.

$x \geq - 1$

Step 3.2

Set the argument in $\arcsin (x)$ less than or equal to $1$ to find where the expression is defined.

$x \leq 1$

Step 3.3

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

$[- 1, 1]$

Step 4

Use each root to create test intervals.

$x < - 1$

$- 1 < x < \frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2} < x < 1$

$x > 1$

Step 5

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

Tap for more steps...

Step 5.1

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 5.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 5.1.2

Replace $x$ with $- 4$ in the original inequality.

$\arcsin (- 4) > \frac{π}{3}$

Step 5.1.3

The left side $N a N$ is not greater than the right side $1.04719755$ , which means that the given statement is false.

False

Step 5.2

Test a value on the interval $- 1 < x < \frac{\sqrt{3}}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 5.2.1

Choose a value on the interval $- 1 < x < \frac{\sqrt{3}}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 5.2.2

Replace $x$ with $0$ in the original inequality.

$\arcsin (0) > \frac{π}{3}$

Step 5.2.3

The left side $0$ is not greater than the right side $1.04719755$ , which means that the given statement is false.

False

Step 5.3

Test a value on the interval $\frac{\sqrt{3}}{2} < x < 1$ to see if it makes the inequality true.

Tap for more steps...

Step 5.3.1

Choose a value on the interval $\frac{\sqrt{3}}{2} < x < 1$ and see if this value makes the original inequality true.

$x = 0.93$

Step 5.3.2

Replace $x$ with $0.93$ in the original inequality.

$\arcsin (0.93) > \frac{π}{3}$

Step 5.3.3

The left side $1.19441284$ is greater than the right side $1.04719755$ , which means that the given statement is always true.

True

Step 5.4

Test a value on the interval $x > 1$ to see if it makes the inequality true.

Tap for more steps...

Step 5.4.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 5.4.2

Replace $x$ with $4$ in the original inequality.

$\arcsin (4) > \frac{π}{3}$

Step 5.4.3

The left side $N a N$ is not greater than the right side $1.04719755$ , which means that the given statement is false.

False

Step 5.5

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

$x < - 1$ False

$- 1 < x < \frac{\sqrt{3}}{2}$ False

$\frac{\sqrt{3}}{2} < x < 1$ True

$x > 1$ False

$x < - 1$ False

$- 1 < x < \frac{\sqrt{3}}{2}$ False

$\frac{\sqrt{3}}{2} < x < 1$ True

$x > 1$ False

Step 6

The solution consists of all of the true intervals.

$\frac{\sqrt{3}}{2} < x < 1$

Step 7

Convert the inequality to interval notation.

$(\frac{\sqrt{3}}{2}, 1)$

Step 8