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2 cos 3x - 1 = 0
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2cos 3x=1 Work it like any other equation
cos (3x) = 1/2 We know that cos(x) =1/2 when x = Pi/6, so it must be Pi/18
How did you get this? Im still a bit confused :(
Solve 2cos(3x)-1=0 for cos(3x) this gives you: cos(3x)=1/2 as software said previously, we know that cos(x)=1/2 @ x=pi/6 Using this x=pi/6, replace x with 3x: 3x=pi/6 (solve for x) Voila, x=pi/18; depending on your interval (ie 0<x<pi)
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