Y=2sinx-cosx on [0, 2pi]
Find x- and y-intercepts
y=2sin(x)-cos(x)
so when x=0
y=2sin(0)-cos(0)
y=0-1
y=-1
and when y=0
0=2sin(x)-cos(x)
cos(x)=2sin(x)
cos(x)/sin(x)=2
cot(x)=2
x=cot^-1(2)
x=26.5651 which is not between 0 and 2 pi
To find the x-intercepts, we need to solve the equation when y is equal to zero.
Setting y to zero, we have:
0 = 2sin(x) - cos(x)
To solve this equation, we can rearrange it as follows:
cos(x) = 2sin(x)
Now, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) in terms of cos^2(x):
1 - cos^2(x) + cos^2(x) = 1
1 = 1
This identity tells us that the equation 1 = 1 is always true. Therefore, any value of x would satisfy the equation. So, there are infinitely many x-intercepts, and the graph of y = 2sin(x) - cos(x) intersects the x-axis at every multiple of 2π.
To find the y-intercept, we can substitute x = 0 into the equation y = 2sin(x) - cos(x):
y = 2sin(0) - cos(0)
y = 0 - 1
y = -1
Therefore, the y-intercept of the graph is -1.