Please help with equation involving inverse trigonometric functions

Solve for x:

arccos x +2arcsin(√3/2)= π/3

I keep getting 1/2 for my answer but it does not check. Here's my process

arccos x + 2arcsin(√3/2)= π/3

arccos x = π/3 - 2arcsin(√3/2)

x=cos[π/3 - 2arcsin(√3/2)]

Using cosine of a difference identity:

x=cos(π/3)cos2[arcsin(√3/2)] + sin(π/3)sin2[arcsin(√3/2)]

x=(1/2)cos(2π/3) + (√3/2)sin(2π/3)

x=(1/2)(-1/2) + (√3/2)(√3/2)

x= (-1/4) + (3/4)

x=2/4=1/2

Could someone please tell me what I'm doing wrong?

Re: Please help with equation involving inverse trigonometric functions

Quote:

arccos x +2arcsin(√3/2)= π/3

$\displaystyle \arccos{x} + 2 \cdot \frac{\pi}{3} = \frac{\pi}{3}$

$\displaystyle \arccos{x} = -\frac{\pi}{3}$

the range of the $\displaystyle \arccos$ function is $\displaystyle [0,\pi]$ ... this equation has no solution.