# Two trig equations

• Mar 18th 2010, 10:22 AM
ldx2
Two trig equations
I ran into problems with these two trig equations, can someone help me ?

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

$\displaystyle arccosx+arccos(1-x)=arccos(-x)$
• Mar 18th 2010, 11:31 AM
icemanfan
For the first problem:

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

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

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

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

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

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

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

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

Substitute $\displaystyle z = x^2$ yielding

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

and solve the quadratic for z:

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

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

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

Hence

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

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

Using the definition of the arccos and arcsin, the only solution that works is $\displaystyle x = \frac{1}{2}$.
• Mar 18th 2010, 11:55 AM
ldx2
Thank you very much, i understand it now. Do you have any ideas for the second equation ?
• Mar 18th 2010, 12:16 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:

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

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

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

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

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

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

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

$\displaystyle 2x - x^2 = 1 - x^2$

$\displaystyle 2x = 1$

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

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