Trigonometry Examples

\sin^{2} (x) = 0

$\sin^{2} (x) = 0$

Step 1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

\sin (x) = \pm \sqrt{0}

$\sin (x) = \pm \sqrt{0}$

Step 2

Simplify

\pm \sqrt{0}

$\pm \sqrt{0}$ .

Tap for more steps...

Step 2.1

Rewrite

0

$0$ as

0^{2}

$0^{2}$ .

\sin (x) = \pm \sqrt{0^{2}}

$\sin (x) = \pm \sqrt{0^{2}}$

Step 2.2

Pull terms out from under the radical, assuming positive real numbers.

\sin (x) = \pm 0

$\sin (x) = \pm 0$

Step 2.3

Plus or minus

0

$0$ is

0

$0$ .

\sin (x) = 0

$\sin (x) = 0$

\sin (x) = 0

$\sin (x) = 0$

Step 3

Take the inverse sine of both sides of the equation to extract

x

$x$ from inside the sine.

x = \arcsin (0)

$x = \arcsin (0)$

Step 4

Simplify the right side.

Tap for more steps...

Step 4.1

The exact value of

\arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 5

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

π

$π$ to find the solution in the second quadrant.

x = π - 0

$x = π - 0$

Step 6

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 7

Find the period of

\sin (x)

$\sin (x)$ .

Tap for more steps...

Step 7.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 7.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 7.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 7.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 8

The period of the

\sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 2 π n, π + 2 π n

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

n

$n$

Step 9

Consolidate the answers.

x = π n

$x = π n$ , for any integer

n

$n$