# Solving an inverse trig equation

1. Feb 28, 2009

### Kaylee!

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Feb 28, 2009

### 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.