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Solve-for-x-cos-x-7-cos-2x-7-cos-3x-7-1-2-




Question Number 15786 by tawa tawa last updated on 13/Jun/17
Solve for x  cos((x/7)) − cos(((2x)/7)) + cos(((3x)/7)) = (1/2)
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{7}}\right)\:−\:\mathrm{cos}\left(\frac{\mathrm{2x}}{\mathrm{7}}\right)\:+\:\mathrm{cos}\left(\frac{\mathrm{3x}}{\mathrm{7}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by tawa tawa last updated on 14/Jun/17
please help with this.
$$\mathrm{please}\:\mathrm{help}\:\mathrm{with}\:\mathrm{this}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17
(x/7)=t  cost−cos2t+cos3t=(1/2)  cost−(2cos^2 t−1)+(4cos^3 t−3cost)=(1/2)  ⇒8cos^3 t−4cos^2 t−4cost+1=0  ⇒cost=−.62,.22,.9  cos(x/7)=−.62=cos(−θ)⇒x=14kπ±cos^(−1) (.62)  cos(x/7)=.22⇒x=14mπ±cos^(−1) (.22)  cos(x/7)=.9⇒x=14nπ±cos^(−1) (.9)  ⇒x=2rπ±cos^(−1) (.9,.22,.65) .r∈Z
$$\frac{{x}}{\mathrm{7}}={t} \\ $$$${cost}−{cos}\mathrm{2}{t}+{cos}\mathrm{3}{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${cost}−\left(\mathrm{2}{cos}^{\mathrm{2}} {t}−\mathrm{1}\right)+\left(\mathrm{4}{cos}^{\mathrm{3}} {t}−\mathrm{3}{cost}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{8}{cos}^{\mathrm{3}} {t}−\mathrm{4}{cos}^{\mathrm{2}} {t}−\mathrm{4}{cost}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{cost}=−.\mathrm{62},.\mathrm{22},.\mathrm{9} \\ $$$${cos}\frac{{x}}{\mathrm{7}}=−.\mathrm{62}={cos}\left(−\theta\right)\Rightarrow{x}=\mathrm{14}{k}\pi\pm{cos}^{−\mathrm{1}} \left(.\mathrm{62}\right) \\ $$$${cos}\frac{{x}}{\mathrm{7}}=.\mathrm{22}\Rightarrow{x}=\mathrm{14}{m}\pi\pm{cos}^{−\mathrm{1}} \left(.\mathrm{22}\right) \\ $$$${cos}\frac{{x}}{\mathrm{7}}=.\mathrm{9}\Rightarrow{x}=\mathrm{14}{n}\pi\pm{cos}^{−\mathrm{1}} \left(.\mathrm{9}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}{r}\pi\pm{cos}^{−\mathrm{1}} \left(.\mathrm{9},.\mathrm{22},.\mathrm{65}\right)\:.{r}\in{Z} \\ $$
Commented by tawa tawa last updated on 14/Jun/17
God bless you sir. i really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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