Applying the trigonometric identity: 2cos(θ)2−1=cos(2θ)2\cos\left(\theta \right)^2-1 = \cos\left(2\theta \right)2cos(θ)2−1=cos(2θ)
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8cos2x−2cosx=18cos^2x-2cosx=18cos2x−2cosx=1
x2−3x−10x2−4\frac{x^2-3x-10}{x^2-4}x2−4x2−3x−10
4x2−3−3x+4−234x^2-3-\sqrt{3}x+4-2\sqrt{3}4x2−3−3x+4−23
((3x+1)cos(2x))e(2x) \frac{\left(\left(3x+1\right)cos\left(2x\right)\right)}{e^{\left(2x\right)}}\:e(2x)((3x+1)cos(2x))
x2−x3−2x4x^2-x^3-2x^4x2−x3−2x4
−1x−18-1x-18−1x−18
ln(x−2)=ln(2)+ln(x−3)\ln\left(x-2\right)=\ln\left(2\right)+\ln\left(x-3\right)ln(x−2)=ln(2)+ln(x−3)
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