Math Help - Trig: Equations with Inverse Trigonometric Functions?

1. Trig: Equations with Inverse Trigonometric Functions?

I need help with solving this problem:

arccos x + 2 arcsin (√3/2) = pi/3

I keep getting 1/2, and when I plug it back into the equation it equals pi instead of pi/3. Here's what I did:

arccos x + 2 arcsin (√3/2) = pi/3

arccos x = pi/3 - 2 arcsin (√3/2)

arccos x = -pi/3

x = cos (-pi/3)

x = 1/2

What did I do wrong? What is the correct answer?

I really need help now cuz my math test is Monday! D:

2. looks to me that you did everything correct.

the range of $y = \arccos{x}$ is $0 \leq y \leq \pi$

as I see it, this equation has no solution.

3. I would've done exactly the same as you did there . Odd.