Question about SoftMath Algebrator - Algebra Homework Solver (689076614429)

# Cos2x + 3 = 5cosx

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Posted by Anonymous on

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Use the identity cos(2x)=2(cos(x))^2-1
cos(2x)+3=5cos(x) becomes 2(cos(x))^2-1+3=5cos(x)
Arrange a bit: 2(cos(x))^2-5cos(x)+2=0
Get rid of the 2-factor
(cos(x))^2-(5/2) cos(x)+1=0
This is a quadratic equation for the unknown U=cos(x)
U^2-(5/2)U+1=0
Solve it by factoring or with the quadratic equation formula. The solutions are U=2 or U=1/2.
Since U=cos(x), the root U=cox(x)=2 must be rejected.
What is left is cos(x)=(1/2). The solutions are x=60 or x=-60 plus or minus 360 degrees.

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### Cos2x + 3 = 5cosx

the solution you got is correct it is

cos X =2 and cos X =1/2

but we know maximum value of cosX is 1.so we discard the solution cos X =2

so only solution is cos X=1/2

and X = 60 degrees

hope you r satisfied.please rate the solution high.thank you

Apr 17, 2010 | SoftMath Algebrator - Algebra Homework...

### Everytime i try to differentiate or intergrate it never replies with an answer, only empty brackets "{ }". Say i want to differentiate (cos2x) how would i input it, because d(cos(2*x),x) isnt working...

DelVar x , I think on your calculator x might be part of a list so, delete it at the main folder should work. If you hit NewProb before you start it will delete all single varibles for you so, you won't have conflicts using them. You have everything correct, its just your ti memory that missed you up.

If for some reason this doesn't work, let know by adding a comment and I'll quickly help you so, you can get on and won't have hang ups.

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