# Solve algebraically and graphically: $\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$

Solve:

$$\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$$

I think the algebraic solution should start like:

$$\arcsin(x) = \frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)$$ $$x = \sin\left(\frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)\right)$$

at that stage probably I should use some trigonometric relation or property of $\arccos(x)$ to convert it to $\arcsin(x)$, however I can't figure it out.

As for the solution by graph I can't even think what steps should I follow to build it.

$$\arcsin x + \arccos\frac{x}{2} = \frac{5\pi}{6}$$
$$\arcsin x = \frac{5\pi}{6} - \arccos\frac{x}{2}$$ $$x = \sin{\left(\frac{5\pi}{6} - \arccos\frac{x}{2}\right)}$$ $$x=\sin\frac{5\pi}6\cos \left(\arccos\frac{x}{2}\right)-\cos\frac{5\pi}6 \sin\left(\arccos\frac{x}{2}\right)$$ $$x=\frac12\cdot\frac x2+\frac{\sqrt3}{2} \cdot\sqrt{1-\left(\frac x2\right)^2}$$ Answer: $x=1$
First you have the sine of one thing minus another. $$\sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta.$$ After applying that, you will have $$\sin\arccos w = \sqrt{1-w^2} \tag 1$$ $$\cos\arccos w = w$$ To get the identity $(1)$, just draw a triangle. If $\cos\alpha =w$ then you have $\text{adjacent} = w$ and $\text{hypotenuse} = 1$, so by the Pythagorean theorem you have $\text{opposite} = \sqrt{1-w^2}.$ Then recall that $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}}.$
HINT: $$\arccos\dfrac x2-\dfrac\pi3=\dfrac\pi2-\arcsin x=\arccos x$$
Applying cosine $$\dfrac x2\cdot\cos\dfrac\pi3+\sin\dfrac\pi3\sqrt{1-\left(\dfrac x2\right)^2}=x$$
$$\implies x=1$$