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Question Number 100988 by bobhans last updated on 29/Jun/20
4sin^2 x + sin 2x = 3 find solution set on x∈(0,2π)
$$\mathrm{4sin}\:^{\mathrm{2}} {x}\:+\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{3}\: \\ $$$${find}\:{solution}\:{set}\:{on}\:{x}\in\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$
Commented by Dwaipayan Shikari last updated on 29/Jun/20
sin^2 x+2sinxcosx+cos^2 x=3−3sin^2 x+cos^2 x (sinx+cosx)^2 =4cos^2 x sinx+cosx=2cosx sinx=cosx sinx=sin((π/2)−x) x=kπ+(−1)^k ((π/2)−x) 2x=kπ+(π/2) 4x=2kπ+π x=(2k+1)(π/4) {k∈Z so solution x∈[0,2π] are (π/4),((3π)/4) ,((5π)/4),((7π)/4) but at x=((3π)/4) ,((7π)/4) are not valid It has another generic solution x=kπ−((3π)/4) net set∈{(π/4),(π/2)+tan^(−1) (1/3),((5π)/4),((3π)/2)+tan^(−1) (1/3)}
$${sin}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx}+{cos}^{\mathrm{2}} {x}=\mathrm{3}−\mathrm{3}{sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x} \\ $$$$\:\left({sinx}+{cosx}\right)^{\mathrm{2}} =\mathrm{4}{cos}^{\mathrm{2}} {x} \\ $$$$\:\:{sinx}+{cosx}=\mathrm{2}{cosx} \\ $$$$\:\:{sinx}={cosx} \\ $$$$\:\:{sinx}={sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\:\:{x}={k}\pi+\left(−\mathrm{1}\right)^{{k}} \left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$\:\:\mathrm{2}{x}={k}\pi+\frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{4}{x}=\mathrm{2}{k}\pi+\pi \\ $$$$\:\:{x}=\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{4}}\:\:\left\{{k}\in\mathbb{Z}\right. \\ $$$$\:\:{so}\:{solution}\:\:{x}\in\left[\mathrm{0},\mathrm{2}\pi\right]\:\:{are}\:\frac{\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{4}}\:,\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{7}\pi}{\mathrm{4}}\:\:{but}\:{at}\:\:{x}=\frac{\mathrm{3}\pi}{\mathrm{4}}\:\:,\frac{\mathrm{7}\pi}{\mathrm{4}}\:\:{are}\:{not}\:{valid} \\ $$$$ \\ $$$${It}\:{has}\:{another}\:{generic}\:{solution}\:{x}={k}\pi−\frac{\mathrm{3}\pi}{\mathrm{4}} \\ $$$${net}\:{set}\in\left\{\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{4}},\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$
Answered by MJS last updated on 29/Jun/20
4sin^2 x +sin 2x =3 t=tan x ((t^2 +2t−3)/(t^2 +1))=0 t_1 =−3 ⇒ x_1 =arctan (1/3) +(n−(1/2))π t_2 =1 ⇒ x_2 =(π/4)+nπ 0≤x<2π ⇒ x∈{(π/4), (π/2)+arctan (1/3), ((5π)/4), ((3π)/2)+arctan (1/3)}
$$\mathrm{4sin}^{\mathrm{2}} \:{x}\:+\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{3} \\ $$$${t}=\mathrm{tan}\:{x} \\ $$$$\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{3}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{0} \\ $$$${t}_{\mathrm{1}} =−\mathrm{3}\:\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\:+\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi \\ $$$${t}_{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{2}} =\frac{\pi}{\mathrm{4}}+{n}\pi \\ $$$$\mathrm{0}\leqslant{x}<\mathrm{2}\pi\:\Rightarrow\:{x}\in\left\{\frac{\pi}{\mathrm{4}},\:\frac{\pi}{\mathrm{2}}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}},\:\frac{\mathrm{5}\pi}{\mathrm{4}},\:\frac{\mathrm{3}\pi}{\mathrm{2}}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{3}}\right\} \\ $$

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