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Question Number 199424 by cortano12 last updated on 03/Nov/23

$$\:\:\:\boldsymbol{{x}} \\ $$

Answered by Frix last updated on 04/Nov/23

f(x)=((cos x)/3)(6sin^3 x −4sin^2 x +1) f′(x)=−sin x (1−2sin x)(3−4sin^2 x) The rest is easy. I solved it for 0≤x<2π f(0)=(1/3) (max) f((π/6))=((√3)/8) (min) f((π/3))=−((8−9(√3))/(24)) (max) f(((2π)/3))=((8−9(√3))/(24)) (min) f(((5π)/6))=−((√3)/8) (max) f(π)=−(1/3) (min) f(((4π)/3))=((8+9(√3))/(24)) (max) f(((5π)/3))=−((8+9(√3))/(24)) (min)

$${f}\left({x}\right)=\frac{\mathrm{cos}\:{x}}{\mathrm{3}}\left(\mathrm{6sin}^{\mathrm{3}} \:{x}\:−\mathrm{4sin}^{\mathrm{2}} \:{x}\:+\mathrm{1}\right) \\ $$$${f}'\left({x}\right)=−\mathrm{sin}\:{x}\:\left(\mathrm{1}−\mathrm{2sin}\:{x}\right)\left(\mathrm{3}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right) \\ $$$$\mathrm{The}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy}. \\ $$$$\mathrm{I}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{for}\:\mathrm{0}\leqslant{x}<\mathrm{2}\pi \\ $$$${f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{3}}\:\left(\mathrm{max}\right) \\ $$$${f}\left(\frac{\pi}{\mathrm{6}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\:\left(\mathrm{min}\right) \\ $$$${f}\left(\frac{\pi}{\mathrm{3}}\right)=−\frac{\mathrm{8}−\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{24}}\:\left(\mathrm{max}\right) \\ $$$${f}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\frac{\mathrm{8}−\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{24}}\:\left(\mathrm{min}\right) \\ $$$${f}\left(\frac{\mathrm{5}\pi}{\mathrm{6}}\right)=−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}\:\left(\mathrm{max}\right) \\ $$$${f}\left(\pi\right)=−\frac{\mathrm{1}}{\mathrm{3}}\:\left(\mathrm{min}\right) \\ $$$${f}\left(\frac{\mathrm{4}\pi}{\mathrm{3}}\right)=\frac{\mathrm{8}+\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{24}}\:\left(\mathrm{max}\right) \\ $$$${f}\left(\frac{\mathrm{5}\pi}{\mathrm{3}}\right)=−\frac{\mathrm{8}+\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{24}}\:\left(\mathrm{min}\right) \\ $$

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