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Question Number 30054 by rahul 19 last updated on 15/Feb/18

Answered by ajfour last updated on 16/Feb/18

let  sin x= s  5s^2 +4s^2 (1−s^2 )−4(1−2s^2 ) > 0  or   4s^4 −17s^2 +4 < 0    (4s^2 −1)(s^2 −4) < 0  ⇒  (1/4)< s^2  < 4   and we know s^2  ≤1  so   (1/4)< s^2  ≤ 1  ⇒ n𝛑+(𝛑/6) < x < (n+1)𝛑−(𝛑/6)  .

$${let}\:\:\mathrm{sin}\:{x}=\:{s} \\ $$$$\mathrm{5}{s}^{\mathrm{2}} +\mathrm{4}{s}^{\mathrm{2}} \left(\mathrm{1}−{s}^{\mathrm{2}} \right)−\mathrm{4}\left(\mathrm{1}−\mathrm{2}{s}^{\mathrm{2}} \right)\:>\:\mathrm{0} \\ $$$${or}\:\:\:\mathrm{4}\boldsymbol{{s}}^{\mathrm{4}} −\mathrm{17}\boldsymbol{{s}}^{\mathrm{2}} +\mathrm{4}\:<\:\mathrm{0} \\ $$$$\:\:\left(\mathrm{4}\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{1}\right)\left(\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{4}\right)\:<\:\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{4}}<\:{s}^{\mathrm{2}} \:<\:\mathrm{4}\:\:\:{and}\:{we}\:{know}\:{s}^{\mathrm{2}} \:\leqslant\mathrm{1} \\ $$$${so}\:\:\:\frac{\mathrm{1}}{\mathrm{4}}<\:{s}^{\mathrm{2}} \:\leqslant\:\mathrm{1} \\ $$$$\Rightarrow\:\boldsymbol{{n}\pi}+\frac{\boldsymbol{\pi}}{\mathrm{6}}\:<\:\boldsymbol{{x}}\:<\:\left(\boldsymbol{{n}}+\mathrm{1}\right)\boldsymbol{\pi}−\frac{\boldsymbol{\pi}}{\mathrm{6}}\:\:. \\ $$

Commented by rahul 19 last updated on 16/Feb/18

thank u sir.

$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir}. \\ $$

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