Solving an inverse trig equation

[Kaylee!]

Homework Statement

Solve for x
arcsin(2x) + arccos(x) = \frac{\pi}{6}

Homework Equations

The Attempt at a Solution

Subtracting arccos(x) from both sides and then taking the sine of both sides:
2x = sin\left[\frac{\pi}{6} - arccos(x)\right]

Applying the difference angle identity for sine:
2x = sin \frac{\pi}{6} cos[arccos (x)] - cos \frac{\pi}{6} sin [arccos (x)]

We can draw a right triangle with a hypotenuse of 1, angle opposite of theta of \sqrt{1-x^2}, and angle adjacent to theta of x. From this triangle we can deduce the values of cos( arccos(x) ) and sin( arccos(x) )
2x = (\frac{1}{2})x - \frac{\sqrt{3}}{2}{\sqrt{1-x^2}}

Multiplying each side by 2, subtracting x from each side, squaring each side, and then dividing each side by 3
3x^2 = 1 - x^2

Adding x^2 to each side, and then dividing each side by 4
x^2 = \frac{1}{4}

At this point I need to take the square root of each side. I know it's the negative from trying each out, but how do I show this algebraically?

 

[csprof2000]

When you say 2x = x/2 - sqrt(3)/2 sqrt(1-x^2)...

3x = -sqrt(3)sqrt(1-x^2).

But since sqrt(1-x^2) is positive, the RHS is negative, which is true iff the LHS is negative, which is true iff x is negative, so x must be negative.