I have to solve the equation:
$$2 \arcsin x = \arccos 2x$$
I see that we have the following conditions:
$$\hspace{6cm} x \in [-1, 1] \hspace{5cm} (1)$$
from the left-hand side and
$$2x \in [-1, 1]$$
$$\hspace{6cm} x \in \bigg [-\dfrac{1}{2}, \dfrac{1}{2} \bigg ] \hspace{5cm} (2)$$
from the right-hand side. Combining $(1)$ and $(2)$, we have that:
$$x \in \bigg [ – \dfrac{1}{2}, \dfrac{1}{2} \bigg ]$$
But I don't know how to solve the equation itself. I tried something like this:
$$\sin (2 \arcsin x) = \sin (\arccos 2x)$$
And something like this:
$$\cos(2 \arcsin x) = \cos (\arccos 2x)$$
$$\cos (2 \arcsin x) = 2x$$
But either way, I don't know how to continue.
Best Answer
$\cos (2t)=1-2(\sin t)^{2}$. So $\cos (2\arcsin x)=2x$ is same as $1-2x^{2}=2x$. Solve this quadratic to find $x$.