\(\quad \cos\left(6x\right)\\ =\cos(2\cdot3x)\\ =2\cos^2(3x)-1\\ \text{Know: }\cos(3x) = 4\cos^3(x) - 3 \cos (x)\\ =2\left(4\cos^3(x)-3\cos(x)\right)^2 - 1\\ =2\left(16\cos^6(x)-24\cos^4(x)+9\cos^2(x)\right)-1\\ =32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1\)
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