x^3 - 3x = sqrt(x+2).

hint - try trigo in this.

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 !!!