Solve-2sin-2x-3-sin-x-cos-x-2-0-

Question Number 1870 by alib last updated on 18/Oct/15

$${Solve}\: \\ $$$$ \\ $$$$\mathrm{2}{sin}\:\mathrm{2}{x}−\:\mathrm{3}\left({sin}\:{x}\:+\:{cos}\:{x}\right)\:+\:\mathrm{2}\:=\mathrm{0} \\ $$

Answered by 112358 last updated on 19/Oct/15

2sin2x−3(sinx+cosx)+2=0 4sinxcosx−3sinx−3cosx+2=0 2sin^2 x+4sinxcosx+2cos^2 x=3(sinx+cosx) 2(sinx+cosx)^2 −3(sinx+cosx)=0 (sinx+cosx)(2sinx+2cosx−3)=0 We obtain that sinx+cosx=0 or 2(sinx+cosx)−3=0 (1) sinx+cosx=0 ∵ sinx=tanxcosx ⇒ tanxcosx+cosx=0 cosx(tanx+1)=0 ⇒cosx=0⇔x=2nπ±(π/2) n∈Z or tanx=−1⇔x=mπ−(π/4) m∈Z Checking the solution x=0.5π 2sinπ−3(sin0.5π)−3cos0.5π+2=−1≠0 ∴x≠2nπ±0.5π Define S_1 as solution set 1. ∴ S_1 ={x∈R, m∈N∣ x=mπ−(π/4)} (2) 2(sinx+cosx)−3=0 In harmonic form sinx+cosx=(√2)sin(x+(π/4)) ∴2(√2)sin(x+(π/4))=3 sin(x+(π/4))=(3/(2(√2))) But, (3/2)=1.5>(√2)=1.41...⇒(3/(2(√2)))>1. Since ∣sinp∣≤1 ∀p∈R,⇒sin(x+(π/4))=(3/(2(√2))) has no real solution. The only set of real solutions is S_1 ={x∈R∣ x=mπ−(π/4),m∈N}

$$\mathrm{2}{sin}\mathrm{2}{x}−\mathrm{3}\left({sinx}+{cosx}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{sinxcosx}−\mathrm{3}{sinx}−\mathrm{3}{cosx}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}{sin}^{\mathrm{2}} {x}+\mathrm{4}{sinxcosx}+\mathrm{2}{cos}^{\mathrm{2}} {x}=\mathrm{3}\left({sinx}+{cosx}\right) \\ $$$$\mathrm{2}\left({sinx}+{cosx}\right)^{\mathrm{2}} −\mathrm{3}\left({sinx}+{cosx}\right)=\mathrm{0} \\ $$$$\left({sinx}+{cosx}\right)\left(\mathrm{2}{sinx}+\mathrm{2}{cosx}−\mathrm{3}\right)=\mathrm{0} \\ $$$${We}\:{obtain}\:{that}\: \\ $$$${sinx}+{cosx}=\mathrm{0}\:\:{or}\:\:\mathrm{2}\left({sinx}+{cosx}\right)−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:{sinx}+{cosx}=\mathrm{0} \\ $$$$\because\:{sinx}={tanxcosx} \\ $$$$\Rightarrow\:{tanxcosx}+{cosx}=\mathrm{0} \\ $$$${cosx}\left({tanx}+\mathrm{1}\right)=\mathrm{0}\: \\ $$$$\Rightarrow{cosx}=\mathrm{0}\Leftrightarrow{x}=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}}\:\:{n}\in\mathbb{Z} \\ $$$${or}\:{tanx}=−\mathrm{1}\Leftrightarrow{x}={m}\pi−\frac{\pi}{\mathrm{4}}\:\:{m}\in\mathbb{Z} \\ $$$${Checking}\:{the}\:{solution}\:{x}=\mathrm{0}.\mathrm{5}\pi \\ $$$$\mathrm{2}{sin}\pi−\mathrm{3}\left({sin}\mathrm{0}.\mathrm{5}\pi\right)−\mathrm{3}{cos}\mathrm{0}.\mathrm{5}\pi+\mathrm{2}=−\mathrm{1}\neq\mathrm{0} \\ $$$$\therefore{x}\neq\mathrm{2}{n}\pi\pm\mathrm{0}.\mathrm{5}\pi\: \\ $$$${Define}\:{S}_{\mathrm{1}} \:{as}\:{solution}\:{set}\:\mathrm{1}. \\ $$$$\therefore\:{S}_{\mathrm{1}} =\left\{{x}\in\mathbb{R},\:{m}\in\mathbb{N}\mid\:{x}={m}\pi−\frac{\pi}{\mathrm{4}}\right\} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}\left({sinx}+{cosx}\right)−\mathrm{3}=\mathrm{0} \\ $$$${In}\:{harmonic}\:{form} \\ $$$${sinx}+{cosx}=\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\therefore\mathrm{2}\sqrt{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{3} \\ $$$${sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${But},\:\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5}>\sqrt{\mathrm{2}}=\mathrm{1}.\mathrm{41}…\Rightarrow\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}}>\mathrm{1}. \\ $$$${Since}\:\mid{sinp}\mid\leqslant\mathrm{1}\:\forall{p}\in\mathbb{R},\Rightarrow{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$${has}\:{no}\:{real}\:{solution}. \\ $$$${The}\:{only}\:{set}\:{of}\:{real}\:{solutions}\:{is} \\ $$$${S}_{\mathrm{1}} =\left\{{x}\in\mathbb{R}\mid\:{x}={m}\pi−\frac{\pi}{\mathrm{4}},{m}\in\mathbb{N}\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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