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Question Number 130047 by liberty last updated on 22/Jan/21

Given 4cos^2 x sin x−2sin^2 x = 3sin x where −(π/2)≤x≤(π/2) . Find the sum of all posibble value of x.

$$\:\mathrm{Given}\:\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{sin}\:\mathrm{x}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{3sin}\:\mathrm{x} \\ $$$$\mathrm{where}\:−\frac{\pi}{\mathrm{2}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of} \\ $$$$\mathrm{all}\:\mathrm{posibble}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$

Answered by MJS_new last updated on 22/Jan/21

sin x =s ⇒ cos^2 x =1−s^2 (s^2 +(s/2)−(1/4))s=0 ⇒ s=0∨s=−(1/4)±((√5)/4) ⇒ x=0∨x=−((3π)/(10))∨x=(π/(10)) ⇒ answer is −(π/5)

$$\mathrm{sin}\:{x}\:={s}\:\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\mathrm{1}−{s}^{\mathrm{2}} \\ $$$$\left({s}^{\mathrm{2}} +\frac{{s}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}\right){s}=\mathrm{0} \\ $$$$\Rightarrow\:{s}=\mathrm{0}\vee{s}=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$$\Rightarrow\:{x}=\mathrm{0}\vee{x}=−\frac{\mathrm{3}\pi}{\mathrm{10}}\vee{x}=\frac{\pi}{\mathrm{10}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\frac{\pi}{\mathrm{5}} \\ $$

Answered by Alepro3 last updated on 22/Jan/21

cos^2 x=1−sin^2 x ⇒ 4sin x−4sin^3 x−2sin^2 x=3sin x ⇒ sin x(4sin^2 x+2sin x−sin x)=0 ⇒ one solution is sin x=0 ⇒ x=kπ solving the 2^(nd) degree equation we have sin x_(1,2) =(−2±(√(4+16)))/8=(−1±(√5))/4 so x_(1,2) =sin^(−1) [(−1±(√5))/4]+2kπ and x_(3,4) =π−sin^(−1) [[(1−(√5))/4]+2kπ ⇒ the sum of all of this value of x is Σx=π+5kπ

$$ \\ $$$$\mathrm{cos}\:^{\mathrm{2}} {x}=\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\:\:\Rightarrow \\ $$$$\mathrm{4sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}−\mathrm{2sin}\:^{\mathrm{2}} {x}=\mathrm{3sin}\:{x}\:\:\Rightarrow \\ $$$$\mathrm{sin}\:{x}\left(\mathrm{4sin}\:^{\mathrm{2}} {x}+\mathrm{2sin}\:{x}−\mathrm{sin}\:{x}\right)=\mathrm{0}\:\:\Rightarrow \\ $$$${one}\:{solution}\:{is}\:\mathrm{sin}\:{x}=\mathrm{0}\:\:\Rightarrow\:\:{x}={k}\pi \\ $$$${solving}\:{the}\:\mathrm{2}^{{nd}} {degree}\:{equation}\:{we}\:{have} \\ $$$$\mathrm{sin}\:{x}_{\mathrm{1},\mathrm{2}} =\left(−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{16}}\right)/\mathrm{8}=\left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\right)/\mathrm{4} \\ $$$${so}\:{x}_{\mathrm{1},\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \left[\left(−\mathrm{1}\pm\sqrt{\mathrm{5}}\right)/\mathrm{4}\right]+\mathrm{2}{k}\pi\:\:{and}\:\:{x}_{\mathrm{3},\mathrm{4}} =\pi−\mathrm{sin}^{−\mathrm{1}} \left[\left[\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)/\mathrm{4}\right]+\mathrm{2}{k}\pi\:\:\Rightarrow\right. \\ $$$${the}\:{sum}\:{of}\:{all}\:{of}\:{this}\:{value}\:{of}\:{x}\:{is} \\ $$$$\Sigma{x}=\pi+\mathrm{5}{k}\pi \\ $$

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