(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Solve for x

[tex]arcsin(2x) + arccos(x) = \frac{\pi}{6}[/tex]

2. Relevant equations

3. The attempt at a solution

Subtracting arccos(x) from both sides and then taking the sine of both sides:

[tex]2x = sin\left[\frac{\pi}{6} - arccos(x)\right][/tex]

Applying the difference angle identity for sine:

[tex]2x = sin \frac{\pi}{6} cos[arccos (x)] - cos \frac{\pi}{6} sin [arccos (x)] [/tex]

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

[tex]2x = (\frac{1}{2})x - \frac{\sqrt{3}}{2}{\sqrt{1-x^2}}[/tex]

Multiplying each side by 2, subtracting x from each side, squaring each side, and then dividing each side by 3

[tex]3x^2 = 1 - x^2[/tex]

Adding x^2 to each side, and then dividing each side by 4

[tex]x^2 = \frac{1}{4}[/tex]

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?

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# Solving an inverse trig equation

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