Question

Please help.
is there a way to find the four solutions for 1/2 + 1/2cos(2x) =a
given that 0 11 years ago

Answer 1

If you pick a value for x lets say pi/4
Then you can find a value for a which would be 1/2+1/2cos(2*pi/4) = a
So then a = 1/2 and that works,
So now you just have to do that for more values of x and those give you your values of a. You just need to make sure a is between 0 and 1

Answer 2

is there more to it? more info?

Answer 3

looks pretty ambiguous to me

Answer 4

@Arfney thanks, what can I do after cause when we move the1/2 to it the equation become 0

Answer 5

@jdoe0001 nope,that is all I got T T

Answer 6

hmm

Answer 7

In my method you "guess" x and solve for A. Then if A is between 0 and 1 then you have a solution.

Answer 8

@Arfney thanks again :)

Answer 9

Your welcome! :D

Answer 10

\(\bf \cfrac{1}{2}+\cfrac{1}{2}cos(2x)=a\implies \cfrac{1}{2}+\cfrac{cos(2x)}{2}=a\implies \cfrac{1+cos(2x)}{2}=a
\\ \quad \\
1+cos(2x)=2a\implies \cancel{ 1 }+[{\color{brown}{ 2cos^2(x)\cancel{ -1}}}]=2a
\\ \quad \\
2cos^2(x)=2a\implies cos^2(x)=a\implies cos(x)=\pm \sqrt{a}
\\ \quad \\
x=cos^{-1}(\pm \sqrt{a})\)

is thus far what I have

Answer 11

but I can see many "a" between 0 and 1, having an angle

Answer 12

@jdoe0001 yeah I can see many a too :( and thanks so much

Answer 13

hmmm unless you're expected some rational number..... I gather...

as Arfney said, if you use say 1/4 for that then \(\bf x=cos^{-1}\left(\pm \sqrt{\frac{1}{4}}\right)\implies x=cos^{-1}\left(\pm \sqrt{\frac{1^2}{2^2}}\right)\implies x=cos^{-1}\left(\pm \frac{1}{2}\right)\)

and that should give you an angle, and then I guess you could use the reference angle of that for the other 3 quadrants

so that angle in each quadrant, would give 4 angles that are between \(\bf 0..2\pi\)