Solve the equation $$\cos\dfrac{2x}{3}+\sqrt3\sin\dfrac{2x}{3}=\dfrac{8}{3+\cos{4x}}$$
Let's divide both sides of the equation by $2$ to get $$\dfrac12\cos\dfrac{2x}{3}+\dfrac{\sqrt3}{2}\sin\dfrac{2x}{3}=\dfrac{4}{3+\cos{4x}}\\\iff \sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=\dfrac{4}{3+\cos4x}$$ The RHS appears to be $\ge1$. Then isn't then the equation equivalent to $$\sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=1$$ which gives $x=\dfrac{\pi}{2}+3k\pi,k\in\mathbb{Z}$?
Why in the authors' solution is the equation said to be equivalent to the system $$\begin{cases}\sin\left(\dfrac{\pi}{6}+\dfrac{2x}{3}\right)=1\\\cos{4x}=1\end{cases}$$ and then it's said that all of the solutions of the first equation satisfy the second. Why is the second necessary at all?
Best Answer
To begin with, I would recommend you to notice that \begin{align*} -1 \leq \cos(4x) \leq 1 & \Longleftrightarrow 2 \leq \cos(4x) + 3 \leq 4\\\\ & \Longleftrightarrow \frac{1}{4} \leq \frac{1}{\cos(4x) + 3} \leq \frac{1}{2}\\\\ & \Longleftrightarrow 1 \leq \frac{4}{\cos(4x) + 3} \leq 2 \end{align*}
Hence the LHS equals the RHS iff one has that \begin{align*} 1 \leq \frac{4}{\cos(4x) + 3} = \sin\left(\frac{2x}{3} + \frac{\pi}{6}\right) \leq 1 \end{align*}
Hopefully this helps!