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Algebra
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i will explain all the things i tried out with this problem.
first i tried the rational root theorem.
x^3 - 3x = sqrt(x+2).
(x^3-3x)^2 = x + 2.
by the rational root theorem we can check that + 2 or -2 can be a root.
plugging it in the initial equation we get x = 2 as a solution. x = -2 doesn't work.
but this was not the whole solution.
then see that range for real values of x is [-2,infinity).
clearly cause x + 2 > or = 0.
now this part was tricky.
for all x > 2, x(x^-4) > 0.
also x^2 > x + 2 for x > 2
since (x-2)(x+1) > 0 for all x > 2,
so x > sqrt(x+2). ...(1)
but we just proved that x^3 - 4x > 0.
or, x^3 - 3x > x.... (2).
combining (1) and (2) we get LHS > RHS for all x > 2.
so we zero in to the possible range of x that is [-2,2].
now comes trigo !!
put x = 2cosb. cause cosb lies in [-1,1] so, 2cosb lies in [-2,2].
we have to simplify
8cos^3b - 6cosb = sqrt(2(cosb+1))
cosb +1 = 2cos^2(b/2).
also LHS becomes 2cos3b.
so we have 2cos3b = sqrt(4cos^2(b/2))
or, cos3b = cos(b/2).
3b = 2n(pi) + or - (b/2).
notice here that b lies in [0,pi]. cause then b/2 lies in [0,pi/2] and is therefore +ve.
2n(pi) = 7b/2 or 5b/2.
we put n = 0 to get b = 0 or, x = 2cos0 = 2.
we put n = 1 to get b = 4pi/5 or b = 4pi/7.
hence x = 2cos(4pi/7)
x = 2cos(4pi/5).
the question is done.
and please don't ask for complex solutions for x !!!
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