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Question Number 105112 by bemath last updated on 26/Jul/20

2(√3) sin^2 (x+((3π)/2)) + sin 2x = 0 0≤x≤2π

$$\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:^{\mathrm{2}} \left({x}+\frac{\mathrm{3}\pi}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\mathrm{0}\: \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\: \\ $$

Answered by john santu last updated on 28/Jul/20

⇔ 2(√3) cos^2 (x)+ sin (2x) = 0 2cos (x) {(√3) cos x+sin x } = 0 { ((cos x = 0 →x = (π/2)+2kπ)),((tan x = −(√3) = tan ((2π)/3) →x=((2π)/3)+kπ)) :} { ((x = (π/2))),((x = ((2π)/3) ; ((5π)/3))) :} (JS ♠★)

$$\Leftrightarrow\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:^{\mathrm{2}} \left({x}\right)+\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{2cos}\:\left({x}\right)\:\left\{\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\:\right\}\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{cos}\:{x}\:=\:\mathrm{0}\:\rightarrow{x}\:=\:\frac{\pi}{\mathrm{2}}+\mathrm{2}{k}\pi}\\{\mathrm{tan}\:{x}\:=\:−\sqrt{\mathrm{3}}\:=\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:\rightarrow{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}+{k}\pi}\end{cases} \\ $$$$\begin{cases}{{x}\:=\:\frac{\pi}{\mathrm{2}}}\\{{x}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:;\:\frac{\mathrm{5}\pi}{\mathrm{3}}}\end{cases}\:\:\:\left({JS}\:\spadesuit\bigstar\right) \\ $$

Answered by Dwaipayan Shikari last updated on 26/Jul/20

2(√3)cos^2 x+2sinxcosx=0 4cosx(((√3)/2)cosx+(1/2)sinx)=0 4cosx=0 x=(4k+1)(π/2)......(a) or(((√3)/2)cosx+(1/2)sinx)=0 sin((π/3)+x)=0 (π/3)+x=kπ x=kπ±(π/3) { ((x=(π/2))),((x=((2π)/3),((5π)/3))) :}

$$\mathrm{2}\sqrt{\mathrm{3}}{cos}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx}=\mathrm{0} \\ $$$$\mathrm{4}{cosx}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosx}+\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)=\mathrm{0} \\ $$$$\mathrm{4}{cosx}=\mathrm{0} \\ $$$${x}=\left(\mathrm{4}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}......\left({a}\right) \\ $$$${or}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosx}+\frac{\mathrm{1}}{\mathrm{2}}{sinx}\right)=\mathrm{0} \\ $$$${sin}\left(\frac{\pi}{\mathrm{3}}+{x}\right)=\mathrm{0} \\ $$$$\frac{\pi}{\mathrm{3}}+{x}={k}\pi \\ $$$${x}={k}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$$\begin{cases}{{x}=\frac{\pi}{\mathrm{2}}}\\{{x}=\frac{\mathrm{2}\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{3}}}\end{cases} \\ $$

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