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Solving an inverse trig equation

  1. Feb 28, 2009 #1
    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?
     
  2. jcsd
  3. Feb 28, 2009 #2
    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.
     
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