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.

 

[Architeuthis Dux]

Great job on solving the inverse trig equation! You are on the right track. To show the negative solution algebraically, you can take the square root of both sides and then use the "plus-minus" symbol, denoted as ±, to represent both the positive and negative solutions. So the final solution would be ±x = ±\frac{1}{2}. This means that there are actually four possible solutions for x: x = \frac{1}{2}, x = -\frac{1}{2}, x = -\frac{1}{2}, or x = \frac{1}{2}. You can plug these values back into the original equation to check which ones satisfy the equation. Keep up the good work!