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sin^2(x)+cos(2x)-cos(x)=0?
How do I tackle this? What identities should I use? At first glance I want to get rid of that sin^2(x) to get something with a cosine in it, but I don't know how to proceed.
2 Answers
- Anonymous9 years agoFavorite Answer
sin^2(x)+cos(2x)-cos(x)=0
As cos(2x)=cos^2(x)-sin^2(x) then:
sin^2(x)+cos^2(x)-sin^2(x)-cos(x)=0
cos^2(x)-cos(x)=0
cos(x)(cos(x)-1)=0
cos(x)=0 and cos(x)=1
When cos(x)=0 x1=+-pi/2+2*pi*n
When cos(x)=1 x=2*pi*n
Source(s): http://eduboard.com/math/trigonometry/ - MelvynLv 79 years ago
1- cos^2 x + 2Cos2 x -1 -c os x =0
Cos^2 x- cos[x]=0
(Cos x)(cos x -1)=0
cos x=0 -> x = Pi/2 +/- n Pi
cos x=1-> x== +/- n Pi
