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
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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