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Question Number 155989 by cortano last updated on 07/Oct/21

    5sin^2 2x + 8cos^3 x = 8cos x      ((3π)/2)≤x≤2π

$$\:\:\:\:\mathrm{5sin}\:^{\mathrm{2}} \mathrm{2x}\:+\:\mathrm{8cos}\:^{\mathrm{3}} \mathrm{x}\:=\:\mathrm{8cos}\:\mathrm{x} \\ $$$$\:\:\:\:\frac{\mathrm{3}\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$

Commented by john_santu last updated on 07/Oct/21

 x=((3π)/2), 2π−arccos (2/3), 2π

$$\:{x}=\frac{\mathrm{3}\pi}{\mathrm{2}},\:\mathrm{2}\pi−\mathrm{arccos}\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{2}\pi \\ $$

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