Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 4429 by alib last updated on 24/Jan/16

(sin x +(√3) cos x) sin 3x = 2

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

Answered by Yozzii last updated on 24/Jan/16

sinx+(√3)cosx=(√(1+3))sin(x+π/3) =2sin(x+(π/3)) (sinx+(√3)cosx)sin3x=2.........(Υ) ∴ 2sin(x+π/3)sin3x=2 sin(x+π/3)sin3x=1 −0.5(cos(4x+π/3)−cos(2x−π/3))=1 cos(4x+π/3)−cos(2x−π/3)=−2 (∗) (∗) is true iff cos(4x+π/3)=−1 and cos(2x−π/3)=1. From cos(4x+π/3)=−1⇒4x+π/3=2nπ±π (n∈Z) 4x=(2n−(1/3)±1)π x=((6n−1±3)/(12))π⇒ x=(((3n−2)π)/6) or x=(((3n+1)π)/6). From cos(2x−π/3)=1 ⇒2x−π/3=2rπ⇒x=(1/2)(2rπ+(1/3))=(((6r+1)π)/6)=(r+(1/6))π (r∈Z) Let X_1 ={x∈R,n∈Z∣x=(((3n−2)π)/6)}, X_2 ={x∈R,n∈Z∣x=(((3n+1)π)/6)}, X_3 ={x∈R,r∈Z∣x=(((6r+1)π)/6)}. These are the three solution sets obtained above. q may or may not equal n. We must search for a set of values of x that satisfies both equations. So, we investigate when X_1 =X_3 and when X_2 =X_(3 ) . If X_1 =X_3 ⇒3n−2=6r+1⇒3(n−2r)=3⇒n=2r+1 ∴ n=2r+1 (odd n). So for (Υ), sin3x=sin(3(((π(6r+1))/6))) =sin(π(((6r+1)/2))) =sin(3πr+(π/2)) =sin3πrcos(0.5π)+cos3πrsin0.5π sin3x=(−1)^r 2sin(x+(π/3))=2sin(πr+(π/6)+(π/3)) =2sin(πr+((3+6)/(18))π) =2sin(πr+0.5π) =2(sinπrcos0.5π+sin0.5πcosπr) =2(−1)^r ∴ 2(−1)^r (−1)^r =2(−1)^(2r) =2 X_3 can be used on its own as one general solution set covering X_1 . If X_2 =X_3 ⇒3n+1=6r+1⇒n=2r (even n). Now, Z=O∪E. Since n∈Z⇒ n∈O or n∈E , O and E being disjoint. Since X_1 =X_3 for odd n and X_2 =X_3 for even n, for all r∈Z in each case, then X_3 =X_1 ∪X_2 ; i.e X_3 constitutes all solutions of (Υ).

$${sinx}+\sqrt{\mathrm{3}}{cosx}=\sqrt{\mathrm{1}+\mathrm{3}}{sin}\left({x}+\pi/\mathrm{3}\right) \\ $$$$=\mathrm{2}{sin}\left({x}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$\left({sinx}+\sqrt{\mathrm{3}}{cosx}\right){sin}\mathrm{3}{x}=\mathrm{2}.........\left(\Upsilon\right) \\ $$$$\therefore\:\mathrm{2}{sin}\left({x}+\pi/\mathrm{3}\right){sin}\mathrm{3}{x}=\mathrm{2} \\ $$$${sin}\left({x}+\pi/\mathrm{3}\right){sin}\mathrm{3}{x}=\mathrm{1}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$−\mathrm{0}.\mathrm{5}\left({cos}\left(\mathrm{4}{x}+\pi/\mathrm{3}\right)−{cos}\left(\mathrm{2}{x}−\pi/\mathrm{3}\right)\right)=\mathrm{1} \\ $$$${cos}\left(\mathrm{4}{x}+\pi/\mathrm{3}\right)−{cos}\left(\mathrm{2}{x}−\pi/\mathrm{3}\right)=−\mathrm{2}\:\:\:\left(\ast\right) \\ $$$$\left(\ast\right)\:{is}\:{true}\:{iff}\:{cos}\left(\mathrm{4}{x}+\pi/\mathrm{3}\right)=−\mathrm{1} \\ $$$${and}\:{cos}\left(\mathrm{2}{x}−\pi/\mathrm{3}\right)=\mathrm{1}. \\ $$$${From}\:{cos}\left(\mathrm{4}{x}+\pi/\mathrm{3}\right)=−\mathrm{1}\Rightarrow\mathrm{4}{x}+\pi/\mathrm{3}=\mathrm{2}{n}\pi\pm\pi\:\:\:\left({n}\in\mathbb{Z}\right) \\ $$$$\mathrm{4}{x}=\left(\mathrm{2}{n}−\frac{\mathrm{1}}{\mathrm{3}}\pm\mathrm{1}\right)\pi \\ $$$${x}=\frac{\mathrm{6}{n}−\mathrm{1}\pm\mathrm{3}}{\mathrm{12}}\pi\Rightarrow\:{x}=\frac{\left(\mathrm{3}{n}−\mathrm{2}\right)\pi}{\mathrm{6}}\:\:{or}\:{x}=\frac{\left(\mathrm{3}{n}+\mathrm{1}\right)\pi}{\mathrm{6}}. \\ $$$${From}\:{cos}\left(\mathrm{2}{x}−\pi/\mathrm{3}\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}{x}−\pi/\mathrm{3}=\mathrm{2}{r}\pi\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{r}\pi+\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\left(\mathrm{6}{r}+\mathrm{1}\right)\pi}{\mathrm{6}}=\left({r}+\frac{\mathrm{1}}{\mathrm{6}}\right)\pi\:\:\left({r}\in\mathbb{Z}\right) \\ $$$$ \\ $$$${Let}\:{X}_{\mathrm{1}} =\left\{{x}\in\mathbb{R},{n}\in\mathbb{Z}\mid{x}=\frac{\left(\mathrm{3}{n}−\mathrm{2}\right)\pi}{\mathrm{6}}\right\}, \\ $$$${X}_{\mathrm{2}} =\left\{{x}\in\mathbb{R},{n}\in\mathbb{Z}\mid{x}=\frac{\left(\mathrm{3}{n}+\mathrm{1}\right)\pi}{\mathrm{6}}\right\}, \\ $$$${X}_{\mathrm{3}} =\left\{{x}\in\mathbb{R},{r}\in\mathbb{Z}\mid{x}=\frac{\left(\mathrm{6}{r}+\mathrm{1}\right)\pi}{\mathrm{6}}\right\}. \\ $$$${These}\:{are}\:{the}\:{three}\:{solution}\:{sets}\:{obtained} \\ $$$${above}.\:{q}\:{may}\:{or}\:{may}\:{not}\:{equal}\:{n}.\:{We} \\ $$$${must}\:{search}\:{for}\:{a}\:{set}\:{of}\:{values}\:{of}\:{x} \\ $$$${that}\:{satisfies}\:{both}\:{equations}.\:{So},\:{we}\: \\ $$$${investigate}\:{when}\:{X}_{\mathrm{1}} ={X}_{\mathrm{3}} \:{and}\:{when} \\ $$$${X}_{\mathrm{2}} ={X}_{\mathrm{3}\:} . \\ $$$$ \\ $$$${If}\:{X}_{\mathrm{1}} ={X}_{\mathrm{3}} \Rightarrow\mathrm{3}{n}−\mathrm{2}=\mathrm{6}{r}+\mathrm{1}\Rightarrow\mathrm{3}\left({n}−\mathrm{2}{r}\right)=\mathrm{3}\Rightarrow{n}=\mathrm{2}{r}+\mathrm{1} \\ $$$$\therefore\:{n}=\mathrm{2}{r}+\mathrm{1}\:\left({odd}\:{n}\right).\:{So}\:{for}\:\left(\Upsilon\right), \\ $$$${sin}\mathrm{3}{x}={sin}\left(\mathrm{3}\left(\frac{\pi\left(\mathrm{6}{r}+\mathrm{1}\right)}{\mathrm{6}}\right)\right) \\ $$$$={sin}\left(\pi\left(\frac{\mathrm{6}{r}+\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$$={sin}\left(\mathrm{3}\pi{r}+\frac{\pi}{\mathrm{2}}\right) \\ $$$$={sin}\mathrm{3}\pi{rcos}\left(\mathrm{0}.\mathrm{5}\pi\right)+{cos}\mathrm{3}\pi{rsin}\mathrm{0}.\mathrm{5}\pi \\ $$$${sin}\mathrm{3}{x}=\left(−\mathrm{1}\right)^{{r}} \\ $$$$\mathrm{2}{sin}\left({x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{2}{sin}\left(\pi{r}+\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}{sin}\left(\pi{r}+\frac{\mathrm{3}+\mathrm{6}}{\mathrm{18}}\pi\right) \\ $$$$=\mathrm{2}{sin}\left(\pi{r}+\mathrm{0}.\mathrm{5}\pi\right) \\ $$$$=\mathrm{2}\left({sin}\pi{rcos}\mathrm{0}.\mathrm{5}\pi+{sin}\mathrm{0}.\mathrm{5}\pi{cos}\pi{r}\right) \\ $$$$=\mathrm{2}\left(−\mathrm{1}\right)^{{r}} \\ $$$$\therefore\:\mathrm{2}\left(−\mathrm{1}\right)^{{r}} \left(−\mathrm{1}\right)^{{r}} =\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{2}{r}} =\mathrm{2}\:\:\:\: \\ $$$${X}_{\mathrm{3}} \:{can}\:{be}\:{used}\:{on}\:{its}\:{own}\:{as}\:{one}\: \\ $$$${general}\:{solution}\:{set}\:{covering}\:{X}_{\mathrm{1}} . \\ $$$$ \\ $$$${If}\:{X}_{\mathrm{2}} ={X}_{\mathrm{3}} \Rightarrow\mathrm{3}{n}+\mathrm{1}=\mathrm{6}{r}+\mathrm{1}\Rightarrow{n}=\mathrm{2}{r}\:\left({even}\:{n}\right). \\ $$$${Now},\:\mathbb{Z}=\mathbb{O}\cup\mathbb{E}.\:{Since}\:{n}\in\mathbb{Z}\Rightarrow\:{n}\in\mathbb{O}\:{or} \\ $$$${n}\in\mathbb{E}\:,\:\mathbb{O}\:{and}\:\mathbb{E}\:{being}\:{disjoint}.\: \\ $$$${Since}\:{X}_{\mathrm{1}} ={X}_{\mathrm{3}} \:{for}\:{odd}\:{n}\:{and}\:{X}_{\mathrm{2}} ={X}_{\mathrm{3}} \\ $$$${for}\:{even}\:{n},\:{for}\:{all}\:{r}\in\mathbb{Z}\:{in}\:{each}\:{case}, \\ $$$${then}\:{X}_{\mathrm{3}} ={X}_{\mathrm{1}} \cup{X}_{\mathrm{2}} ;\:{i}.{e}\:{X}_{\mathrm{3}} \:{constitutes} \\ $$$${all}\:{solutions}\:{of}\:\left(\Upsilon\right). \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com