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Pianoman1996j
How do you solve \[\cos 2 x= cos x +2\]
do you know cos(x+y) formula?
Yes cosxcosy+sinxsiny
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
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
I think. I will spend some time with it now. Thanks.
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
*typo replace sin^2 with 1-cos^2 : cos^2 - (1 - cos^2)= cos^2 -1 + cos^2 or 2cos^2(x) -1
I think I have it! Hang on while I try to finish!!
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?
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
Of course! Thanks for helping me see.
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
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.
I only had to give solutions between 0 and 2pi but I like your further explaination
Oops. Explanation I mean
Unfortunately, with trig I think you just have to do a bunch of problems and build a sense of what might work.
Lol. I've done hundreds and am still picking the wrong ones. But I'm not giving up yet!
Thanks for help. my midterm is soon. Yikes.