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\[\cos(3 \pi a)+2\cos(2 \pi a)=0\] (Anlge measurents are in radiens ) Prove that \[a=2/3\]
did u try writing cos(3pi*a) using the formula of cos 3x ? and cos(2pi*a) using cos 2x and getting a cubic equation ?
x=pi*a
\[4x^3-4x^2-3x+2=0\]
\[(2x-1)(2x^2-x-2)=0\] \[x=1/2\] \[\cos(3\pi a)=1/2\]
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\[3\pi a=\pm \pi/3+2\pi k\]
i made mistake \[4x^3+4x^3-3x-2=0\] \[(2x+1)(2x^2+x-2)=0\] \[x=-1/2\] \[\cos(\pi a)=\pm (\frac{ 2\pi }{ 3 })+2\pi k\]
last step\[\cos (\pi a)=-1/2\] \[\pi a=\frac{ 2 \pi }{ 3 }\]
yipee
\[a=2/3\]
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is that correct
??
guys can you say something
sorry i was afk, its correct
thanks
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