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 May 14th, 2012, 11:10 PM #1 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 find x ?( (1-?(1-x^2))/2 ) + ?( (1+?(1-x^2))/2 ) = ?(1+x)
 May 15th, 2012, 12:09 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: find x We are given: $\sqrt{\frac{1}{2}$$1-\sqrt{1-x^2}$$}+\sqrt{\frac{1}{2}$$1+\sqrt{1-x^2}$$}=\sqrt{x+1}$ We see we require $|x|\le1$ in order for the left side to represent real values. Multiply through by $\sqrt{2}$: $\sqrt{1-\sqrt{1-x^2}}+\sqrt{1+\sqrt{1-x^2}}=\sqrt{2(x+1)}$ Square both sides since they are non-negative: $1-\sqrt{1-x^2}+2\sqrt{$$1-\sqrt{1-x^2}$$$$1+\sqrt{1-x^2}$$}+1+\sqrt{1-x^2}=2(x+1)$ $2+2\sqrt{1-$$1-x^2$$}=2(x+1)$ $2+2\sqrt{x^2}=2(x+1)$ $2+2|x|=2(x+1)$ Case 1: $0\le x\le 1$ $2+2x=2(x+1)$ $2(x+1)=2(x+1)$ Hence, we find: $0\le x\le1$ Case 2: $-1\le x<0$ $2-2x=2(x+1)$ $x=0$ No solution for this case, thus we are left with: $0\le x\le1$
 May 15th, 2012, 12:35 AM #3 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 Re: find x Thanks a lot!
 May 15th, 2012, 06:13 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find x A sorta "short cut to typing!" I'd use here: since x^2 - 1 = (x+1)(x-1): let a = x+1 and b = x-1; so, after multiplication by SQRT(2), you have: SQRT[1 - SQRT(ab)] + SQRT[1 + SQRT(ab)] = SQRT(2a) Easier to do the squaring both sides et al... I'm lazy
 May 15th, 2012, 09:55 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 Assuming |x| ? 1, let x = sin(2A), where |2A| ? ?/2, and recall that cos(2A) = 2cos²(A) - 1 = 1 - 2sin²(A). The equation becomes |sin(A)| + |cos(A)| = ?(1 + sin(2A)). Squaring gives 1 + |sin(2A)| = 1 + sin(2A), i.e. |x| = x, so 0 ? x ? 1.

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