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If Cosx=0 then what will be the maximum value of x?

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(2n+1)*(pi/2) n=0,1,2,.... Upto infinity
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Here we get different value for x like 90 degree ,270 degree.we calculate up to 360 degree and revise cycle again we get general solution for different n value
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Play with Mathematics

Guys, CosX = 0 = Cos( n.pie +/- 90) Considering maximum value of X n = 1,2,3,.... Infinity So this way Xmax = infinity
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(2n+1)?/2
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90degree
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If Cosx=0 then x=90 degree fix value
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Mathematics classes

general solution is +(2n+1)pi/2 and -(2n+1)pi/2 where n is Integer B'coz integers are unbounded like natural nos. so maximum value cosx = 0 is not exist for general condition if there are some constrains only then get the answer of the question for ex. x belongs to (o,2pi).
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x=(2n+1)*90 where x=integers....
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Experiential Trainer with 8 years of experience in Product Design and Prototyping Industry

90
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Physics,Chemistry,Math

cos90=0 so x=90
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