# Two trig equations

• Mar 18th 2010, 11:22 AM
ldx2
Two trig equations
I ran into problems with these two trig equations, can someone help me ?

$arccosx-arcsinx=arccos\frac{\sqrt{3}}{2}$

$arccosx+arccos(1-x)=arccos(-x)$
• Mar 18th 2010, 12:31 PM
icemanfan
For the first problem:

$\arccos x - \arcsin x = \arccos \frac{\sqrt{3}}{2}$

$\cos (\arccos x - \arcsin x) = \frac{\sqrt{3}}{2}$

$\cos (\arccos x) \cos (\arcsin x) + \sin (\arccos x) \sin (\arcsin x) = \frac{\sqrt{3}}{2}$

$x (\sqrt{1 - x^2}) + x (\sqrt{1 - x^2}) = \frac{\sqrt{3}}{2}$

$2x(\sqrt{1 - x^2}) = \frac{\sqrt{3}}{2}$

$4x^2(1 - x^2) = \frac{3}{4}$

$-4x^4 + 4x^2 = \frac{3}{4}$

$4x^4 - 4x^2 + \frac{3}{4} = 0$

Substitute $z = x^2$ yielding

$16z^2 - 16z + 3 = 0$

and solve the quadratic for z:

$z = \frac{16 \pm \sqrt{256 - 4(16)(3)}}{32}$

$z = \frac{16 \pm \sqrt{64}}{32}$

$z = \frac{16 \pm 8}{32} = \frac{1}{4}, \frac{3}{4}$

Hence

$x^2 = \frac{1}{4}, \frac{3}{4}$

$x = \pm \frac{1}{2}, \pm \frac{\sqrt{3}}{2}$

Using the definition of the arccos and arcsin, the only solution that works is $x = \frac{1}{2}$.
• Mar 18th 2010, 12:55 PM
ldx2
Thank you very much, i understand it now. Do you have any ideas for the second equation ?
• Mar 18th 2010, 01:16 PM
icemanfan
Quote:

Originally Posted by ldx2
Thank you very much, i understand it now. Do you have any ideas for the second equation ?

It's basically the same drill. Take the cosine of both sides:

$\cos(\arccos x + \arccos (1-x)) = -x$

$\cos(\arccos x)\cos(\arccos(1-x)) - \sin(\arccos x)\sin(\arccos(1-x)) = -x$

$x(1-x) - \sqrt{(1-x^2)(2x-x^2)} = -x$

$x(1-x) + x = \sqrt{(1-x^2)(2x-x^2)}$

$(x(1-x) + x)^2 = (1-x^2)(2x-x^2)$

$(2x-x^2)^2 = (1-x^2)(2x-x^2)$

Now x = 2 satisfies this equation. We can look for other solutions by dividing by $2x - x^2$ if $x \neq 2$:

$2x - x^2 = 1 - x^2$

$2x = 1$

$x = \frac{1}{2}$

and since $\arccos 2$ is undefined, the only solution that works is $x = \frac{1}{2}$.