\arccos(1/2) Arccos(1/2) Products
Answer : Note that tan ( x ) = 2 sin ( x ) ⟺ sin ( x ) cos ( x ) = 2 sin ( x ) ⟺ sin ( x ) = 0 ∨ cos ( x ) = 1 2 . \begin{align}\tan(x)=2\sin(x)&\iff\frac{\sin(x)}{\cos(x)}=2\sin(x)\\&\iff\sin(x)=0\vee\cos(x)=\frac12.\end{align} tan ( x ) = 2 sin ( x ) ⟺ cos ( x ) sin ( x ) = 2 sin ( x ) ⟺ sin ( x ) = 0 ∨ cos ( x ) = 2 1 . And cos ( x ) = 1 2 ⟺ x = π 3 + 2 n π ∨ x = − π 3 + 2 n π , \cos(x)=\frac12\iff x=\frac\pi3+2n\pi\vee x=-\frac\pi3+2n\pi, cos ( x ) = 2 1 ⟺ x = 3 π + 2 nπ ∨ x = − 3 π + 2 nπ , with n ∈ Z n\in\Bbb Z n ∈ Z . In particular, although arccos ( 1 2 ) ( = π 3 ) \arccos\left(\frac12\right)\left(=\frac\pi3\right) arccos ( 2 1 ) ( = 3 π ) is indeed a solution of the equation cos ( x ) = 1 2 \cos(x)=\frac12 cos ( x ) = 2 1 , it is not the only one. Let x ∈ [ − π 3 , π 3 ] x\in[-\frac{\pi}{3},\frac{\pi}{3}] x ∈ [ − 3 π , 3 π ] . tan ( x ) = 2 sin ( x ) ⟺ \tan(x)=2\sin(x) \iff tan ( x ) = 2 ...