Find x and y intercepts of y=sinx-cos^2x on
[0, 2pi]
when x=0
y=-1
when y=0
x=0
What is the work for x-intercept? I can't figure it out.
To find the x-intercepts of the given function y = sin(x) - cos^2(x) on the interval [0, 2pi], we need to find the values of x where y = 0.
Setting y = 0, we have:
sin(x) - cos^2(x) = 0
Now, let's solve this equation step-by-step.
Step 1: Rewrite cos^2(x) as 1 - sin^2(x), since cos^2(x) = 1 - sin^2(x).
sin(x) - (1 - sin^2(x)) = 0
Step 2: Expand the equation by removing the parentheses.
sin(x) - 1 + sin^2(x) = 0
Step 3: Rearrange the equation to isolate sin^2(x).
sin^2(x) + sin(x) - 1 = 0
Step 4: Factor the quadratic equation.
(sin(x) - 1)(sin(x) + 1) = 0
Now, we have two possibilities for the factors:
1) sin(x) - 1 = 0
Adding 1 to both sides, we get:
sin(x) = 1
This equation does not have solutions on the interval [0, 2pi].
2) sin(x) + 1 = 0
Subtracting 1 from both sides, we get:
sin(x) = -1
To find the x-intercepts, we need to find the values of x where sin(x) = -1 on the interval [0, 2pi].
The solutions for sin(x) = -1 on this interval are:
x = 3pi/2
Therefore, the x-intercept of the given function on the interval [0, 2pi] is x = 3pi/2.
To find the y-intercept, we need to find the value of y when x = 0.
Substituting x = 0 into the given equation y = sin(x) - cos^2(x), we get:
y = sin(0) - cos^2(0)
y = 0 - 1
y = -1
Therefore, the y-intercept of the given function on the interval [0, 2pi] is y = -1.
In summary:
x-intercept: x = 3pi/2
y-intercept: y = -1
To find the x-intercepts of a function, we need to find the values of x for which y=0. Similarly, to find the y-intercepts, we need to find the values of y when x=0.
Let's find the x-intercepts first:
To find the x-intercepts of the function y = sin(x) - cos^2(x), we set y = 0 and solve for x.
0 = sin(x) - cos^2(x)
Now, let's solve this equation. Rearranging it a bit, we get:
sin(x) = cos^2(x)
Now, using the identity cos^2(x) = 1 - sin^2(x), we can substitute and simplify the equation:
sin(x) = 1 - sin^2(x)
Rearranging this equation, we get:
sin^2(x) + sin(x) - 1 = 0
Now, let's solve this quadratic equation for sin(x). We can factor it or use the quadratic formula:
(sin(x) - 1)(sin(x) + 1) = 0
So either sin(x) - 1 = 0 or sin(x) + 1 = 0.
If sin(x) - 1 = 0, then sin(x) = 1. The sine function has a value of 1 at x = pi/2 and x = 3pi/2. However, we are only interested in the values of x in the interval [0, 2pi].
If sin(x) + 1 = 0, then sin(x) = -1. The sine function has a value of -1 at x = 3pi/2. Again, we are only interested in the values of x in the interval [0, 2pi].
So the x-intercepts of the function y = sin(x) - cos^2(x) in the interval [0, 2pi] are x = pi/2 and x = 3pi/2.
Now, let's find the y-intercept:
To find the y-intercept, we substitute x = 0 into the equation y = sin(x) - cos^2(x):
y = sin(0) - cos^2(0)
= 0 - 1
= -1
So the y-intercept of the function y = sin(x) - cos^2(x) is y = -1.
In summary, the x-intercepts of the function y = sin(x) - cos^2(x) in the interval [0, 2pi] are x = pi/2 and x = 3pi/2, and the y-intercept is y = -1.