How do you solve for the exact solutions in the interval [0,2pi] of cos3x=(1/2)cos3x=(12)?

1 Answer
Bdub
Apr 18, 2016

S={pi/9, (5pi)/9,(7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9 }S={π9,5π9,7π9,11π9,13π9,17π9}

Explanation:

3x=cos^-1 (1/2)3x=cos−1(12)

3x=+- pi/3 +2pin3x=±π3+2πn

x=+- pi/3 xx 1/3+2pin xx 1/3x=±π3×13+2πn×13

x=+-pi/9+(2pi)/3 nx=±π9+2π3n

Now pick n values (positive and negative integers) into the equation

n=0, x=pi/9,-pi/9n=0,x=π9,−π9

n=1, x=(5pi)/9,(7pi)/9n=1,x=5π9,7π9

n=2, x=(11pi)/9, (13pi)/9n=2,x=11π9,13π9

n=3, x=(17pi)/9,(19pi)/9n=3,x=17π9,19π9

S={pi/9, (5pi)/9,(7pi)/9, (11pi)/9, (13pi)/9, (17pi)/9 }S={π9,5π9,7π9,11π9,13π9,17π9}

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