Recall that the sine of an angle in standard position is the second coordinate of the point of intersection of the angle's terminal side and the unit circle. There are two points on the unit circle with second coordinate $0 .$

These points correspond to angles of $θ = 0$ and $θ = π$ radians. These are the solutions to the equation $sin θ = 0 .$

$sin θ = \frac{2}{2}$

To find a solution to this equation, we will use an inverse trigonometric function and a calculator.

sin θ = \frac{2}{2}

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

θ = sin^{- 1} (\frac{2}{2})

UseCalc

Use a calculator

θ = \frac{π}{4}

One solution to

sin θ = \frac{2}{2}

is

θ = \frac{π}{4} .

To find a second solution to this equation, we need to use the unit circle. Remember, sine is positive in the first and second quadrants of the coordinate plane.

The sine of an angle in standard position is the $y -$ coordinate of the point of intersection of the angle's terminal side and the unit circle. The solution $θ = \frac{π}{4}$ is in Quadrant I, so the second solution must be in Quadrant II. Because these angles are symmetric across the $y -$ axis and half a turn measures $π$ radians, to find this angle we subtract $\frac{π}{4}$ from $π .$

Another solution to the equation $sin θ = \frac{2}{2}$ is $θ = \frac{3 π}{4} .$

$sin θ = - \frac{2}{2}$

To find a solution to this equation, we will use an inverse trigonometric function and a calculator.

sin θ = - \frac{2}{2}

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

θ = sin^{- 1} (- \frac{2}{2})

UseCalc

Use a calculator

θ = - \frac{π}{4}

One solution to

sin θ = - \frac{2}{2}

is

θ = - \frac{π}{4} .

To find a second solution to this equation, we need to use the unit circle. Remember, sine is negative in the third and fourth quadrants of the coordinate plane.

The solution $θ = - \frac{π}{4}$ is in Quadrant IV, so the second solution must be in Quadrant III, and these angles are symmetric across the $y -$ axis. Knowing that half a turn measures $π$ radians and a full turn measures $2 π$ radians, we can calculate the desired angle measures. We will add $\frac{π}{4}$ to $π,$ and subtract $\frac{π}{4}$ from $2 π .$

The solutions to the equation $sin θ = - \frac{2}{2}$ are $θ = \frac{5 π}{4}$ and $θ = \frac{7 π}{4} .$

$x -$ intercepts

We found six solutions to the given equation,

θ = 0,

θ = π,

θ = \frac{π}{4},

θ = \frac{3 π}{4},

θ = \frac{5 π}{4},

and

θ = \frac{7 π}{4} .

Keep in mind that if we add or subtract a multiple of

2 π

radians, the terminal side of the angle will be in the same position. Therefore, the resulting angles will also be solutions to the equation.

0 + 2 π n radians, π + 2 π n radians, \frac{π}{4} + 2 π n radians, \frac{3 π}{4} + 2 π n radians, \frac{5 π}{4} + 2 π n radians, \frac{7 π}{4} + 2 π n radians, where n is any integer

Note that the first

two

solutions differ by

π,

and

π

is

half

of

2 π .

Also, the other

four

solutions differ from each other by

\frac{π}{2},

and

\frac{π}{2}

is a

quarter

of

2 π .

With this information in mind, we can write our answer in a simpler form.

0 + π n radians, \frac{π}{4} + \frac{π n}{2} radians, where n is any integer

These are the

x -

intercepts of the given function.

sinθ=0

sinθ=22​​

sinθ=-22​​

x-intercepts

$sin θ = 0$

$sin θ = \frac{2}{2}$

$sin θ = - \frac{2}{2}$

$x -$ intercepts

Exercise