How Can You Further Simplify 2cos2x - 2cosx in Trigonometry?

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Homework Statement


2cos2x-2cosx...how do you simplify this further?


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The Attempt at a Solution


2(cos2x-cosx)..but i have to find 0=2cos2x-2cosx so this doesn't really help me.
 
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You're trying to find when 2cos(2x)-2cos(x)=0? Start by what you did... you want to find x such that cos(2x)=cos(x). Hint: cosine is periodic
 
would it be pi/2 and 3pi/2?
 
nope...maybe try drawing cos(2x) and cos(x)?
 
\cos(2x)=2\cos^{2}(x)-1, so 2\cos(2x)-2\cos(x)=2\cos^{2}(x)-1\cos(x)-1=0. Factoring yields (2\cos(x)+1)(\cos(x)-1)=0
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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