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[Post New]27 Jan 2008 21:43:23 IST
    Subject: Re:The number of values of x in the interval [ 0,5]
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akhil_o (2709)

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Olaaa!! Perrrfect answer. 435  [702 rates]
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fn can be factorised as
 
(3sin x-1)(sin x - 2)=0
hence sin x =1/3 ( 2 not poss)
now sin x = 1/3 takes 2 values in [0,pi]
and none from pi to 2pi
rotating 0-> pi 5 times we get
6 solns
 
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