Hello yaykittyeee Originally Posted by

**yaykittyeee** i'm trying to study for a test and i'm having trouble understanding how to solve these, my textbook doesn't really cover this very well. could someone please help me?

solve for x:

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

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

Inverse trig functions are always a pain. Get rid of them as soon as you can. Here's the first one:

Let $\displaystyle y = \arcsin\left(\frac{x}{2}\right)$. So $\displaystyle \sin y = \frac{x}{2}$

Let $\displaystyle z = \arccos\left(\frac{-x\sqrt3}{2}\right)$. So $\displaystyle \cos z = \frac{-x\sqrt3}{2}$

And the original equation becomes: $\displaystyle y+z=\pi$

$\displaystyle \Rightarrow z = \pi -y$

$\displaystyle \Rightarrow \cos z = -\frac{x\sqrt3}{2}= \cos(\pi-y) = -\cos y $

But $\displaystyle \cos y = \sqrt{1 - \sin^2 y}=\sqrt{1-\frac{x^2}{4}}$

So $\displaystyle \frac{x\sqrt3}{2}=\sqrt{1-\frac{x^2}{4}}$

Solve this for $\displaystyle x$ by squaring both sides. You get a quadratic, one of whose roots is not valid. The other is the answer to the question (which is $\displaystyle x = 1$).

Number 2 is a bit trickier, but start it in the same way. Let: $\displaystyle y = \arcsin\left(\frac{2}{x}\right)$ i.e. $\displaystyle \sin y = ...$

$\displaystyle z = \arccos\left(\frac{\sqrt3}{x}\right)$ i.e. $\displaystyle \cos z = ...$

$\displaystyle w = \arccos\left(\frac{1}{x}\right)$ i.e. $\displaystyle \cos w = ...$

Then say:$\displaystyle y-z=w$

and:$\displaystyle \frac{1}{x}=\cos w = \cos(y-z)$$\displaystyle =\cos y \cos z + \sin y \sin z$

$\displaystyle = ...$ (substitute to get everything in terms of $\displaystyle x$)

Then solve the equation in $\displaystyle x$.

Do you want to have another go now?

Grandad