Answer:
12.57
Step-by-step explanation:
I used C=2πr for the formula
the answer is 2
<h2><u>hope</u><u> it</u><u> helps</u></h2>
<u>see</u><u> the</u><u> attachment</u><u> for</u><u> explanation</u>
Answer:
AA
Step-by-step explanation:
The steps are F and H are matched with line marking it congruent and G is vertical and the last is congruent.
Solve cos(4x)-cos(2x)=0 ∀ 0<=x<=2pi ..............(0)
Normal solution:
1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1
cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1)
2. substitute (1) in (0)
2cos^2(2x)-1-cos(2x)=0
3. substitute u=cos(2x)
2u^2-u-1=0
4. Solve for x
factor
(u-1)(u+1/2)=0
=> u=1 or u=-1/2
However, since cos(x) is an even function, so solutions to
{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2)
are all solutions.
5. The cosine function is symmetrical about pi, therefore
cos(-2x)=cos(2*pi-2x),
solution (2) above becomes
{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}
6. Solve each case
cos(2x)=1 => x=0
cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi
cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3
cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3
Summing up,
x={0,pi/3, 2pi/3, pi, 4pi/3}
