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Question Number 58804 by Tawa1 last updated on 30/Apr/19
Find the value of x: 4 cos x + 3 sin x = 2
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{x}:\:\:\:\:\mathrm{4}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:\:=\:\:\mathrm{2} \\ $$
Commented by MJS last updated on 30/Apr/19
generally acos x +bsin x =c x=2arctan t t^2 −((2b)/(a+c))t−((a−c)/(a+c))=0 t=((b±(√(a^2 +b^2 −c^2 )))/(a+c)) x=2πn+2arctan ((b±(√(a^2 +b^2 −c^2 )))/(a+c)); n∈Z even if t∉R
$$\mathrm{generally} \\ $$$${a}\mathrm{cos}\:{x}\:+{b}\mathrm{sin}\:{x}\:={c} \\ $$$${x}=\mathrm{2arctan}\:{t} \\ $$$${t}^{\mathrm{2}} −\frac{\mathrm{2}{b}}{{a}+{c}}{t}−\frac{{a}−{c}}{{a}+{c}}=\mathrm{0} \\ $$$${t}=\frac{{b}\pm\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{a}+{c}} \\ $$$${x}=\mathrm{2}\pi{n}+\mathrm{2arctan}\:\frac{{b}\pm\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{{a}+{c}};\:{n}\in\mathbb{Z} \\ $$$$\mathrm{even}\:\mathrm{if}\:{t}\notin\mathbb{R} \\ $$
Answered by MJS last updated on 30/Apr/19
4cos x +3sin x =2 x=2arctan t 4((1−t^2 )/(1+t^2 ))+3((2t)/(1+t^2 ))=2 t^2 −t−(1/3)=0 t=(1/2)±((√(21))/6) x=2πn+2arctan ((3±(√(21)))/6); n∈Z
$$\mathrm{4cos}\:{x}\:+\mathrm{3sin}\:{x}\:=\mathrm{2} \\ $$$${x}=\mathrm{2arctan}\:{t} \\ $$$$\mathrm{4}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{2} \\ $$$${t}^{\mathrm{2}} −{t}−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{21}}}{\mathrm{6}} \\ $$$${x}=\mathrm{2}\pi{n}+\mathrm{2arctan}\:\frac{\mathrm{3}\pm\sqrt{\mathrm{21}}}{\mathrm{6}};\:{n}\in\mathbb{Z} \\ $$
Commented by Tawa1 last updated on 30/Apr/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 30/Apr/19
Sir, why when n = 1, the equation is not equal to 2
$$\mathrm{Sir},\:\mathrm{why}\:\mathrm{when}\:\:\mathrm{n}\:=\:\mathrm{1},\:\:\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{2} \\ $$
Answered by mr W last updated on 30/Apr/19
4 cos x + 3 sin x = 2 (4/5) cos x + (3/5) sin x = (2/5) cos α cos x + sin α sin x = (2/5) cos (x−α) = (2/5) ⇒x−α=2nπ±cos^(−1) (2/5) ⇒x=2nπ+α±cos^(−1) (2/5) ⇒x=2nπ+cos^(−1) (4/5)±cos^(−1) (2/5)
$$\mathrm{4}\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:\:=\:\:\mathrm{2} \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{cos}\:\mathrm{x}\:+\:\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{sin}\:\mathrm{x}\:\:=\:\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\mathrm{x}\:+\:\mathrm{sin}\:\alpha\:\mathrm{sin}\:\mathrm{x}\:\:=\:\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\left(\mathrm{x}−\alpha\right)\:\:=\:\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{x}−\alpha=\mathrm{2}{n}\pi\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{x}=\mathrm{2}{n}\pi+\alpha\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\Rightarrow{x}=\mathrm{2}{n}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}\pm\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by Tawa1 last updated on 30/Apr/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 30/Apr/19
Sir, when n = 1, the equation is not equal to 2.
$$\mathrm{Sir},\:\mathrm{when}\:\:\mathrm{n}\:=\:\mathrm{1},\:\:\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{2}.\: \\ $$
Commented by mr W last updated on 30/Apr/19
how did you come to that? with n=1: x=2π+cos^(−1) (4/5)+cos^(−1) (2/5) or x=2π+cos^(−1) (4/5)−cos^(−1) (2/5) with x=2π+cos^(−1) (4/5)+cos^(−1) (2/5): cos x=cos (cos^(−1) (4/5)+cos^(−1) (2/5)) =(4/5)×(2/5)−(3/5)×((√(21))/5)=((8−3(√(21)))/(25)) sin x=sin (cos^(−1) (4/5)+cos^(−1) (2/5)) =(3/5)×(2/5)+(4/5)×((√(21))/5)=((6+4(√(21)))/(25)) 4 cos x+3 sin x=((4(8−3(√(21)))+3(6+4(√(21))))/(25)) =((32+18)/(25))=((50)/(25))=2 ! with x=2π+cos^(−1) (4/5)−cos^(−1) (2/5): cos x=cos (cos^(−1) (4/5)−cos^(−1) (2/5)) =(4/5)×(2/5)+(3/5)×((√(21))/5)=((8+3(√(21)))/(25)) sin x=sin (cos^(−1) (4/5)−cos^(−1) (2/5)) =(3/5)×(2/5)−(4/5)×((√(21))/5)=((6−4(√(21)))/(25)) 4 cos x+3 sin x=((4(8+3(√(21)))+3(6−4(√(21))))/(25)) =((32+18)/(25))=((50)/(25))=2 !
$${how}\:{did}\:{you}\:{come}\:{to}\:{that}? \\ $$$$ \\ $$$${with}\:{n}=\mathrm{1}: \\ $$$${x}=\mathrm{2}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$${or} \\ $$$${x}=\mathrm{2}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}} \\ $$$$ \\ $$$${with}\:{x}=\mathrm{2}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}: \\ $$$$\mathrm{cos}\:{x}=\mathrm{cos}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{3}}{\mathrm{5}}×\frac{\sqrt{\mathrm{21}}}{\mathrm{5}}=\frac{\mathrm{8}−\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{25}} \\ $$$$\mathrm{sin}\:{x}=\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{5}}×\frac{\sqrt{\mathrm{21}}}{\mathrm{5}}=\frac{\mathrm{6}+\mathrm{4}\sqrt{\mathrm{21}}}{\mathrm{25}} \\ $$$$\mathrm{4}\:\mathrm{cos}\:{x}+\mathrm{3}\:\mathrm{sin}\:{x}=\frac{\mathrm{4}\left(\mathrm{8}−\mathrm{3}\sqrt{\mathrm{21}}\right)+\mathrm{3}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{21}}\right)}{\mathrm{25}} \\ $$$$=\frac{\mathrm{32}+\mathrm{18}}{\mathrm{25}}=\frac{\mathrm{50}}{\mathrm{25}}=\mathrm{2}\:! \\ $$$$ \\ $$$${with}\:{x}=\mathrm{2}\pi+\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}: \\ $$$$\mathrm{cos}\:{x}=\mathrm{cos}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{5}}×\frac{\sqrt{\mathrm{21}}}{\mathrm{5}}=\frac{\mathrm{8}+\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{25}} \\ $$$$\mathrm{sin}\:{x}=\mathrm{sin}\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{2}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{5}}×\frac{\sqrt{\mathrm{21}}}{\mathrm{5}}=\frac{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{21}}}{\mathrm{25}} \\ $$$$\mathrm{4}\:\mathrm{cos}\:{x}+\mathrm{3}\:\mathrm{sin}\:{x}=\frac{\mathrm{4}\left(\mathrm{8}+\mathrm{3}\sqrt{\mathrm{21}}\right)+\mathrm{3}\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{21}}\right)}{\mathrm{25}} \\ $$$$=\frac{\mathrm{32}+\mathrm{18}}{\mathrm{25}}=\frac{\mathrm{50}}{\mathrm{25}}=\mathrm{2}\:! \\ $$
Commented by Tawa1 last updated on 30/Apr/19
Sir, when i used π = 360, i got 2 now. Before i used π = 3.142
$$\mathrm{Sir},\:\mathrm{when}\:\mathrm{i}\:\mathrm{used}\:\pi\:=\:\mathrm{360},\:\:\mathrm{i}\:\mathrm{got}\:\mathrm{2}\:\mathrm{now}.\:\:\mathrm{Before}\:\mathrm{i}\:\mathrm{used}\:\:\pi\:=\:\mathrm{3}.\mathrm{142} \\ $$
Commented by MJS last updated on 30/Apr/19
maybe on your calculator you must set the mode for the angle usually it′s “deg” for degree (full circle =360°) and “rad” for radiant (full circle =2π)
$$\mathrm{maybe}\:\mathrm{on}\:\mathrm{your}\:\mathrm{calculator}\:\mathrm{you}\:\mathrm{must}\:\mathrm{set}\:\mathrm{the} \\ $$$$\mathrm{mode}\:\mathrm{for}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{usually}\:\mathrm{it}'\mathrm{s}\:“{deg}''\:\mathrm{for}\:\mathrm{degree}\:\left(\mathrm{full}\:\mathrm{circle}\:=\mathrm{360}°\right) \\ $$$$\mathrm{and}\:“{rad}''\:\mathrm{for}\:\mathrm{radiant}\:\left(\mathrm{full}\:\mathrm{circle}\:=\mathrm{2}\pi\right) \\ $$
Commented by Tawa1 last updated on 30/Apr/19
Ohh. Thanks for your time sir
$$\mathrm{Ohh}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$
Commented by MJS last updated on 30/Apr/19
you′re welcome
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$

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