Are 1, 3, 9
...........................
cos (2x) = cos x
2 cos^2 x -1 = cos x using the double angle formula
2 cos ^2 x -cos x -1 =0
factor
(2 cos x+1) ( cos x -1) = 0
using the zero product property
2 cos x+1 =0 cos x -1 =0
2 cos x = -1 cos x =1
cos x = -1/2 cos x=1
taking the arccos of each side
arccos cos x = arccos (-1/2) arccos cos x = arccos 1
x = 120 degrees x=-120 degrees x=0
remember you get 2 values ( 2nd and 3rd quadrant)
these are the principal values
now we need to add 360
x = 120+ 360n x=-120+ 360n x = 0 + 360n where n is an integer
The slopes of two parallel lines must be identical.
We have slope

, so the slope for the parallel line be the same.
Now, to find an equation that also passes through the given point, we use slope-point form,

, where our point

is substituted for

.

Now, we convert to slope-intercept form as such.

And we are done. :) We can verify graphically that these are indeed parallel lines. See attached.
Answer:

Step-by-step explanation:
The Fundamental Theorem of Calculus states that:
![\displaystyle \frac{d}{dx}\left[ \int_a^x f(t)\, dt \right] = f(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft%5B%20%5Cint_a%5Ex%20f%28t%29%5C%2C%20dt%20%20%5Cright%5D%20%3D%20f%28x%29)
Where <em>a</em> is some constant.
We can let:

By substitution:

Taking the derivative of both sides results in:
![\displaystyle g'(s) = \frac{d}{ds}\left[ \int_6^s g(t)\, dt\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20g%27%28s%29%20%3D%20%5Cfrac%7Bd%7D%7Bds%7D%5Cleft%5B%20%5Cint_6%5Es%20g%28t%29%5C%2C%20dt%5Cright%5D)
Hence, by the Fundamental Theorem:
