<s>[INST] Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$ 
Find the simplified form of $\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]$, where $x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]$

My reference gives the solution $\tan^{-1}\frac43-x$, but is it a complete solution ?
My Attempt
Let $\alpha=\cos^{-1}\dfrac{3}{5}\implies \dfrac{3}{5}=\cos\alpha,\;\dfrac{4}{5}=\sin\alpha$
$$
\cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]=\cos^{-1}\bigg[\cos\alpha\cdot\cos x+\sin\alpha\cdot\sin x\bigg]\\
=\cos^{-1}\bigg[\cos\Big(\alpha-x\Big)\bigg]=2n\pi\pm(\alpha-x)=2n\pi\pm\Big(\tan^{-1}\frac{4}{3}-x\Big)\\
=\tan^{-1}\frac{4}{3}-x\quad\text{iff }\tan^{-1}\frac{4}{3}-x\in[0,\pi]
$$
$$
-x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]\quad\&\quad\alpha=\tan^{-1}\frac{4}{3}\in\Big(0,\frac{\pi}{2}\Big)\\
\implies\alpha-x\in\big[\frac{-3\pi}{4},\frac{5\pi}{4}\big]\not\subset[0,\pi]
$$
 [/INST] Your book is wrong. $x\in\Big[-3\pi/4, 3\pi/4\Big]$, which is an interval of length $3\pi/2$. Whatever be the value of $\alpha, \alpha-x$ will belong to an interval of length $3\pi/2$, which means $\alpha-x$ is not confined to $[0, \pi].$
So the answer is $\begin{cases}2\pi-\alpha+x,&x\in[-3\pi/4,\alpha-\pi)\\\alpha-x,&x\in[\alpha-\pi,\alpha]\\x-\alpha,&x\in(\alpha, 3\pi/4]\end{cases}$
</s>