If cosx-2cos2x=0 then x=?
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cosx-2cos2x=0
=>cosx-2(2cos^2 x - 1)=0
=>cosx-4cos^2 x + 2=0
Let cosx be t, then
=>4t^2-t-2=0
=>t=(1+_/33)/8 or (1-_/33)/8
So, cosx=(1+_/33)/8 or (1-_/33)/8
So, x=cos^-1 (1+_/33)/8 or cos^-1 (1-_/33)/8
So, x=32.534° or 126.375°
PLZ MARK IT AS BRAINLIEST..
=>cosx-2(2cos^2 x - 1)=0
=>cosx-4cos^2 x + 2=0
Let cosx be t, then
=>4t^2-t-2=0
=>t=(1+_/33)/8 or (1-_/33)/8
So, cosx=(1+_/33)/8 or (1-_/33)/8
So, x=cos^-1 (1+_/33)/8 or cos^-1 (1-_/33)/8
So, x=32.534° or 126.375°
PLZ MARK IT AS BRAINLIEST..
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