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For the given expression to be maximum , denominator should be minimum which is sin^2x+3sinxcosx+5 cos^2x should be minimum.
let y= sin^2x+3sinxcosx+5cos^2x
2y= 2sin^2x+6sinxcosx+10cos^2x
(I am multiplying by 2 on both sides for simple calculation . )
2y =sin^2x + 2 (sin x) (3cosx) + (3 cos x )^2 + sin^2x+cos^2x
2y= (sin x + 3 cos x ) ^2 +1
The minimum value of (sin x+3 cos x )^2 is 0
since sin x + 3cos x can be expressed as a single sine function,
sin (x+ n ) where n = arcsin ( 3/ sqrt10)
so the minimum value of 2y is 0+1 =1
and that of y is 1/2
so the minimum value of sin^2 x + 3sinxcosx + cos^2x is 0.5
the maximum value of 1/(sin^2 x + 3sinxcosx + cos^2x) is 2
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