find the cuttin points, on the x axis, for graph y= [(4cos2x) - 3]?

Question

how do i get the cutting points i am a bit confused,, i little explanation will help.,,

thanks,,alot!!

Answer 1

The "cuttin" points are the x-intercepts of the graph of a function. We are looking for the x coordinates of the points where the graph of y = 4cos(2x)) - 3 intersects or cuts the x-axis. Points on the x-axis have a y coordinate of zero. So, let y = 0 and solve for the values of x.

Calculations are done in degrees.

0 = 4cos(2x) - 3

3 = 4cos(2x)

(3 / 4) = cos(2x)

The cosine function is equal to 3 / 4 in the first and fourth quadrants, for the same reference angle.

arccos (3 / 4) = 2x or 360 - arccos(3 / 4) = 2x

41.4 = 2x and 360 - 41.4 = 2x

20.7 = x and 180 - 20.7 = x

20.7 = x or 159.3 = x

The period of y = 4cos(2x) - 3 is: period = (360 dgrees) / 2 = 180 degrees.

Answer: 20.7 degrees + (180 degrees)(k) where k is an integer

and

159.3 degrees + (180 degrees)(k) where k is an integer

Answer 2

to know the cutting points on x axis y must be equal to 0

so 4cos2x - 3 = 0
cos 2x = 3/4
2x = 41.41 + 2n(180)
x = 20.7 + n(180)
where is a natural number and n = >0

Answer 3

when it cuts x axis, y=0
so cos2x=3/4
cos2x=cos^2 x-sin^2 x =2cos^2 x -1
so 2cos^2 x=7/4
cos x=sqrt(7/8)
so x = arccos [sqrt(7/8)]

Answer 4

y= [(4cos2x) - 3]
set y = 0
(4cos2x) = 3
cos2x = 0.75
2x = 41.4° + 360n, 318.6° + 360n
x = 20.7° + 180n, 159.3° + 180n