Question

How do you solve
\[\cos 2 x= cos x +2\]

Answer 1

do you know cos(x+y) formula?

Answer 2

Yes cosxcosy+sinxsiny

Answer 3

almost, the add turns into a minus (it's ugly that way)
if you use that with y= x
cos(x+x)= cos^2 x - sin^2 x

replace sin^2 with 1-cos^2 : cos^2 - 1 - cos^2 or
2cos^2(x) -1

Answer 4

you can replace the cos(2x) with 2 cos^2(x) -1
if you call cos(x) y you have
2y^2 -1 = y+2

solve for y

Answer 5

does that make sense?

Answer 6

I think. I will spend some time with it now. Thanks.

Answer 7

of course, after you find y, you find x using
x = inverse cos(y)

because this is a quadratic, you will get two answers for y

then we have to figure out the the answers for every 2pi

Answer 8

*typo

replace sin^2 with 1-cos^2 : cos^2 - (1 - cos^2)= cos^2 -1 + cos^2 or
2cos^2(x) -1

Answer 9

I think I have it! Hang on while I try to finish!!

Answer 10

I got y=3/2 and y=-1. So -1 is at pi but what about 3/2. My teacher says pi is the only answer? Why?

Answer 11

the cosine goes between -1 and +1 (it never goes outside that range) . There is no angle whose cos is > 1 so you can't get 3/2

Answer 12

Of course! Thanks for helping me see.

Answer 13

and because the cosine repeats, you can add any multiple of 2pi to your answer and it will work. to be complete the answer is

\[ \pi + 2\pi n \]where n is any integer

Answer 14

Is there any trick to knowing which formula to try? I often grab one that would be allowed but isn't the one I should have used.

Answer 15

I only had to give solutions between 0 and 2pi but I like your further explaination

Answer 16

Oops. Explanation I mean

Answer 17

Unfortunately, with trig I think you just have to do a bunch of problems and build a sense of what might work.

Answer 18

Lol. I've done hundreds and am still picking the wrong ones. But I'm not giving up yet!

Answer 19

Thanks for help. my midterm is soon. Yikes.