The x-intercepts of the graph of a function occur when y=0. Let's substitute 0 for y into the given equation.
y= 2sin ^4 θ -sin ^2 θ
⇓
0= 2sin ^4 θ -sin ^2 θ
We will start by solving the equation for sin θ . To do so, we will factor out sin ^2 θ and use the Zero Product Property.
To further simplify Equation II, we need to separate the right-hand side into the positive and negative case.
sin θ =± sqrt(1/2)
sin θ =sqrt(1/2)
sin θ =- sqrt(1/2)
sin θ =sqrt(2/4)
sin θ =- sqrt(2/4)
sin θ =sqrt(2)/2
sin θ =- sqrt(2)/2
We obtained, in total, three equations.
lcsin θ =0 & (I) sin θ = sqrt(2)2 & (II) sin θ = - sqrt(2)2 & (III)
Let's solve these equations one at a time.
One solution to sin θ = sqrt(2)2 is θ = π4. 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 = π4 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 π4 from π.
Another solution to the equation sin θ= sqrt(2)2 is θ = 3π4.
sin θ =- sqrt(2)/2
To find a solution to this equation, we will use an inverse trigonometric function and a calculator.
One solution to sin θ =- sqrt(2)2 is θ = - π4. 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 = - π4 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 2π radians, we can calculate the desired angle measures. We will add π4 to π, and subtract π4 from 2π.
The solutions to the equation sin θ=- sqrt(2)2 are θ = 5π4 and θ = 7π4.
x-intercepts
We found six solutions to the given equation, θ = 0, θ = π, θ = π4, θ = 3π4, θ = 5π4, and θ = 7π4. Keep in mind that if we add or subtract a multiple of 2π 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+2π n radians, π+2π n radians, [0.8em]
Ď€/4+2Ď€ n radians, 3Ď€/4+2Ď€ n radians, [0.8em]
5Ď€/4+2Ď€ n radians, 7Ď€/4+2Ď€ n radians, [0.8em]
wheren is any integer
Note that the first two solutions differ by π, and π is half of 2π. Also, the other four solutions differ from each other by π2, and π2 is a quarter of 2π. With this information in mind, we can write our answer in a simpler form.
0+π n radians, π/4 + π n/2 radians, [0.8em] wheren is any integer
These are the x-intercepts of the given function.
