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.
 A: $$\arcsin x + \arccos\frac{x}{2} = \frac{5\pi}{6}$$
$$\arcsin x = \frac{5\pi}{6} - \arccos\frac{x}{2}$$
$$x = \sin{\left(\frac{5\pi}{6} - \arccos\frac{x}{2}\right)}$$
$$x=\sin\frac{5\pi}6\cos \left(\arccos\frac{x}{2}\right)-\cos\frac{5\pi}6 \sin\left(\arccos\frac{x}{2}\right)$$
$$x=\frac12\cdot\frac x2+\frac{\sqrt3}{2} \cdot\sqrt{1-\left(\frac x2\right)^2}$$
Answer: $x=1$
A: 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 $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}}.$
A: HINT:
$$\arccos\dfrac x2-\dfrac\pi3=\dfrac\pi2-\arcsin x=\arccos x$$
Applying cosine   $$\dfrac x2\cdot\cos\dfrac\pi3+\sin\dfrac\pi3\sqrt{1-\left(\dfrac x2\right)^2}=x$$
$$\implies x=1$$
