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Question Number 135868 by liberty last updated on 16/Mar/21

Answered by EDWIN88 last updated on 16/Mar/21

we know that −1≤cos x≤1 for x in the interval  0≤x≤2π ; so we find cos x = ((−x^3 −2x^2 −5x)/2)  then −2≤−x^3 −2x^2 −5x≤2 or   ⇒−2≤x^3 +2x^2 +5x≤2  we get two solution

$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:−\mathrm{1}\leqslant\mathrm{cos}\:\mathrm{x}\leqslant\mathrm{1}\:\mathrm{for}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi\:;\:\mathrm{so}\:\mathrm{we}\:\mathrm{find}\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{−\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}}{\mathrm{2}} \\ $$$$\mathrm{then}\:−\mathrm{2}\leqslant−\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}\leqslant\mathrm{2}\:\mathrm{or}\: \\ $$$$\Rightarrow−\mathrm{2}\leqslant\mathrm{x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}\leqslant\mathrm{2} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{two}\:\mathrm{solution}\: \\ $$

Commented by EDWIN88 last updated on 16/Mar/21

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