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Question Number 49765 by hassentimol last updated on 10/Dec/18

Solve for x in R : 2 × sin(3x+4) + (√( 3 )) = 0

$$\mathrm{Solve}\:\mathrm{for}\:{x}\:\mathrm{in}\:\mathbb{R}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{2}\:×\:\mathrm{sin}\left(\mathrm{3}{x}+\mathrm{4}\right)\:+\:\sqrt{\:\mathrm{3}\:}\:=\:\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 27/Feb/20

(e)⇔sin(3x+4)=−((√3)/2) ⇒sin(3x+4)=sin(−(π/3))⇒ 3x+4=−(π/3)+2kπ or 3x+4 =((4π)/3)+2kπ ⇒3x=−4−(π/3)+2kπ or 3x=−4+((4π)/3) +2kπ ⇒x =−(4/3)−(π/9) +((2kπ)/3) or x=−(4/3)−(π/9) +((2kπ)/3)

$$\left({e}\right)\Leftrightarrow{sin}\left(\mathrm{3}{x}+\mathrm{4}\right)=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{sin}\left(\mathrm{3}{x}+\mathrm{4}\right)={sin}\left(−\frac{\pi}{\mathrm{3}}\right)\Rightarrow \\ $$$$\mathrm{3}{x}+\mathrm{4}=−\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:{or}\:\mathrm{3}{x}+\mathrm{4}\:=\frac{\mathrm{4}\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:\Rightarrow\mathrm{3}{x}=−\mathrm{4}−\frac{\pi}{\mathrm{3}}+\mathrm{2}{k}\pi\:{or} \\ $$$$\mathrm{3}{x}=−\mathrm{4}+\frac{\mathrm{4}\pi}{\mathrm{3}}\:+\mathrm{2}{k}\pi\:\Rightarrow{x}\:=−\frac{\mathrm{4}}{\mathrm{3}}−\frac{\pi}{\mathrm{9}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\:{or}\:{x}=−\frac{\mathrm{4}}{\mathrm{3}}−\frac{\pi}{\mathrm{9}}\:+\frac{\mathrm{2}{k}\pi}{\mathrm{3}} \\ $$

Answered by afachri last updated on 10/Dec/18

2 sin(3x + 4) = −(√3^ ) ⇒ sin(3x + 4) = −((√3^ )/2) consider 3x + 4 = y , so : sin (y) = − sin((π/3)) Sin has negative value in 3^(rd) and 4^(th) quadrant.p •3^(rd) quadrant ⇒ sin(π + y) = −sin ((π/3)) y = π + (π/3) = ((4π)/3) •4^(th) quadrant ⇒ sin(2π − y) = −sin((π/3)) y = 2π − (π/3) = ((5π)/3) ((4π)/3) = 3x + 4 ⇒ x_1 = ((4π − 12)/9) ((5π)/3) = 3x + 4 ⇒ x_2 = ((5π − 12)/9)

$$\mathrm{2}\:\mathrm{sin}\left(\mathrm{3}{x}\:+\:\mathrm{4}\right)\:\:=\:\:−\sqrt{\mathrm{3}^{} }\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{sin}\left(\mathrm{3}{x}\:+\:\mathrm{4}\right)\:=\:−\frac{\sqrt{\mathrm{3}^{} }}{\mathrm{2}} \\ $$$$\mathrm{consider}\:\:\mathrm{3}{x}\:+\:\mathrm{4}\:=\:{y}\:\:,\:\mathrm{so}\::\:\:\:\:\:\:\:\mathrm{sin}\:\:\:\left({y}\right)\:\:\:\:\:\:\:\:\:\:\:=\:\:\:−\:\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{Sin}\:\mathrm{has}\:\mathrm{negative}\:\mathrm{value}\:\mathrm{in}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{and}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{quadrant}.\mathrm{p} \\ $$$$\bullet\mathrm{3}^{\mathrm{rd}} \:\mathrm{quadrant}\:\:\:\Rightarrow\:\:\:\:\:\mathrm{sin}\left(\pi\:\:+\:\:{y}\right)\:\:=\:\:−\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\:\:=\:\:\pi\:+\:\frac{\pi}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{4}\pi}{\mathrm{3}} \\ $$$$\bullet\mathrm{4}^{\mathrm{th}} \:\mathrm{quadrant}\:\:\:\Rightarrow\:\:\:\:\:\mathrm{sin}\left(\mathrm{2}\pi\:−\:{y}\right)\:=\:\:−\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\:\:\:=\:\:\mathrm{2}\pi\:−\:\frac{\pi}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$$$\frac{\mathrm{4}\pi}{\mathrm{3}}\:\:=\:\:\mathrm{3}{x}\:+\:\mathrm{4}\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:{x}_{\mathrm{1}} \:\:=\:\:\frac{\mathrm{4}\pi\:−\:\mathrm{12}}{\mathrm{9}} \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{3}}\:\:=\:\mathrm{3}{x}\:+\:\mathrm{4}\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:{x}_{\mathrm{2}} \:\:=\:\frac{\mathrm{5}\pi\:−\:\mathrm{12}}{\mathrm{9}} \\ $$

Answered by mr W last updated on 10/Dec/18

sin (3x+4)=−((√3)/2) ⇒3x+4=2nπ−(π/3) or (2n+1)π+(π/3) ⇒x=(((6n−1)π)/9)−(4/3) or ((2(3n+2)π)/9)−(4/3) with n=0,±1,±2,...

$$\mathrm{sin}\:\left(\mathrm{3}{x}+\mathrm{4}\right)=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{x}+\mathrm{4}=\mathrm{2}{n}\pi−\frac{\pi}{\mathrm{3}}\:{or}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi+\frac{\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{\left(\mathrm{6}{n}−\mathrm{1}\right)\pi}{\mathrm{9}}−\frac{\mathrm{4}}{\mathrm{3}}\:{or}\:\frac{\mathrm{2}\left(\mathrm{3}{n}+\mathrm{2}\right)\pi}{\mathrm{9}}−\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${with}\:{n}=\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},... \\ $$

Commented by afachri last updated on 10/Dec/18

did i solve wrong sir ??? please let me know Sir :) hope you don′t mind

$$\mathrm{did}\:\mathrm{i}\:\mathrm{solve}\:\mathrm{wrong}\:\mathrm{sir}\:???\:\mathrm{please}\:\mathrm{let}\:\mathrm{me}\:\mathrm{know} \\ $$$$\left.\mathrm{Sir}\::\right)\:\mathrm{hope}\:\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{mind} \\ $$

Commented by mr W last updated on 10/Dec/18

your solution is not complete sir. you take only 3x+4=π+(π/3), but it can also be 3π+(π/3), 5π+(π/3), etc. −π+(π/3),−3π+(π/3), etc. you take only 3x+4=2π−(π/3), but it can also be 4π−(π/3), 6π−(π/3), etc. −(π/3),−2π+(π/3), etc. as the question says x∈R, i.e. −∞<x<+∞.

$${your}\:{solution}\:{is}\:{not}\:{complete}\:{sir}. \\ $$$${you}\:{take}\:{only}\:\mathrm{3}{x}+\mathrm{4}=\pi+\frac{\pi}{\mathrm{3}},\:{but}\:{it}\:{can}\:{also} \\ $$$${be}\:\mathrm{3}\pi+\frac{\pi}{\mathrm{3}},\:\mathrm{5}\pi+\frac{\pi}{\mathrm{3}},\:{etc}.\:−\pi+\frac{\pi}{\mathrm{3}},−\mathrm{3}\pi+\frac{\pi}{\mathrm{3}},\:{etc}. \\ $$$${you}\:{take}\:{only}\:\mathrm{3}{x}+\mathrm{4}=\mathrm{2}\pi−\frac{\pi}{\mathrm{3}},\:{but}\:{it}\:{can}\:{also} \\ $$$${be}\:\mathrm{4}\pi−\frac{\pi}{\mathrm{3}},\:\mathrm{6}\pi−\frac{\pi}{\mathrm{3}},\:{etc}.\:−\frac{\pi}{\mathrm{3}},−\mathrm{2}\pi+\frac{\pi}{\mathrm{3}},\:{etc}. \\ $$$$ \\ $$$${as}\:{the}\:{question}\:{says}\:{x}\in{R},\:{i}.{e}.\:−\infty<{x}<+\infty. \\ $$

Commented by hassentimol last updated on 10/Dec/18

Sir, at the 2nd line, isn′t it 2nπ−((2π)/3) instead for the second one... ?

$$\mathrm{Sir},\:\mathrm{at}\:\mathrm{the}\:\mathrm{2nd}\:\mathrm{line},\:\mathrm{isn}'\mathrm{t}\:\mathrm{it}\:\mathrm{2}{n}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}\:\mathrm{instead} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{one}...\:? \\ $$

Commented by afachri last updated on 10/Dec/18

thanks a lot for your correction and explanation Mr.W , appreciate that. i′ll fix my job again. hope you never be bored Sir. :):)

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot}\:\mathrm{for}\:\mathrm{your}\:\mathrm{correction}\:\mathrm{and}\:\mathrm{explanation} \\ $$$$\mathrm{Mr}.\mathrm{W}\:\:,\:\mathrm{appreciate}\:\mathrm{that}.\:\mathrm{i}'\mathrm{ll}\:\mathrm{fix}\:\mathrm{my}\:\mathrm{job}\:\mathrm{again}. \\ $$$$\left.\mathrm{h}\left.\mathrm{ope}\:\mathrm{you}\:\mathrm{never}\:\mathrm{be}\:\mathrm{bored}\:\mathrm{Sir}.\::\right):\right) \\ $$$$ \\ $$

Commented by mr W last updated on 10/Dec/18

2nπ−((2π)/3) is the same as (2n+1)π+(π/3) because n is any integer here.

$$\mathrm{2}{n}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}\:{is}\:{the}\:{same}\:{as}\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi+\frac{\pi}{\mathrm{3}} \\ $$$${because}\:{n}\:{is}\:{any}\:{integer}\:{here}. \\ $$

Commented by hassentimol last updated on 10/Dec/18

Ok... thanks sir I′ve forgotten it...

$$\mathrm{Ok}...\:\mathrm{thanks}\:\mathrm{sir}\:\mathrm{I}'\mathrm{ve}\:\mathrm{forgotten}\:\mathrm{it}... \\ $$

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