1prophet sir as usual "bang on target" [1]
though squaring is oder way...but u will hav to square two times...substitution one is
more elegant
solve for x
16x+30\sqrt{1-x^2}=17\sqrt{1+x}+17\sqrt{1-x}
where 0<x<1
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7 Answers
62if we square..
256x2+900(1-x2)+960√1-x2=289(1+x+1-x)+289*4√1-x2
900-644x2+960√1-x2=578+1156√1-x2
644-322-644x2+=196√1-x2
644(1-x2) - 322=196√1-x2
This is a quadratic in √1-x2
If i have made any calculation mistakes.. then you can correct them.. but I hope this gives us the desired results [1]
1yes sir u made a mistake in squaring in 1st step :P
waisey this isnt a doubt..
341Let x = \cos 2 \theta; \theta \in \left[0, \frac{\pi}{2} \right]
Then the equation is 16 \cos 2 \theta + 30 sin 2 \theta = 17 \sqrt 2 (\cos \theta + \sin \theta)
which using well known manipulations is equivalent to
\cos ( 2\theta - \alpha) = \cos \left(\theta - \frac{\pi}{4} \right) where \alpha = \cos^{-1} \frac{8}{17}
One solution is given by 2\theta - \alpha = \theta - \frac{\pi}{4} \Rightarrow x= \cos 2\theta = \frac{240}{289}
The other one needs more number crunching than i have time for
6ya excluding some mistakes in calculation the methid by nishant sir is rite!!
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