Simplified form of $\cos^{-1}\big[\frac{3}{5}\cdot\cos x+\frac{4}{5}\cdot\sin x\big]$, where $x\in\big[\frac{-3\pi}{4},\frac{3\pi}{4}\big]$

Question

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 $\alpha=\cos^{-1}\dfrac{3}{5}\implies \dfrac{3}{5}=\cos\alpha,\;\dfrac{4}{5}=\sin\alpha$ $$ \cos^{-1}\bigg[\dfrac{3}{5}\cdot\cos x+\dfrac{4}{5}\cdot\sin x\bigg]=\cos^{-1}\bigg[\cos\alpha\cdot\cos x+\sin\alpha\cdot\sin x\bigg]\\ =\cos^{-1}\bigg[\cos\Big(\alpha-x\Big)\bigg]=2n\pi\pm(\alpha-x)=2n\pi\pm\Big(\tan^{-1}\frac{4}{3}-x\Big)\\ =\tan^{-1}\frac{4}{3}-x\quad\text{iff }\tan^{-1}\frac{4}{3}-x\in[0,\pi] $$ $$ -x\in\Big[\dfrac{-3\pi}{4},\dfrac{3\pi}{4}\Big]\quad\&\quad\alpha=\tan^{-1}\frac{4}{3}\in\Big(0,\frac{\pi}{2}\Big)\\ \implies\alpha-x\in\big[\frac{-3\pi}{4},\frac{5\pi}{4}\big]\not\subset[0,\pi] $$

Answer 1

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 $\begin{cases}2\pi-\alpha+x,&x\in[-3\pi/4,\alpha-\pi)\\\alpha-x,&x\in[\alpha-\pi,\alpha]\\x-\alpha,&x\in(\alpha, 3\pi/4]\end{cases}$

Answer 2

$$-\dfrac{3\pi}4\le x\le\dfrac{3\pi}4$$

$$\iff-\dfrac{3\pi}4-\cos^{-1}\dfrac35\le x-\cos^{-1}\dfrac35\le\dfrac{3\pi}4-\cos^{-1}\dfrac35$$

Now $\dfrac{3\pi}4-\cos^{-1}\dfrac35\le\pi$ as $\cos^{-1}\dfrac35>0>\dfrac{3\pi}4-\pi$

So, $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=x-\cos^{-1}\dfrac35$ if $x-\cos^{-1}\dfrac35\ge0\iff x\ge\cos^{-1}\dfrac35$

Again we can prove $-2\pi<-\dfrac{3\pi}4-\cos^{-1}\dfrac35<-\pi$

For $-\pi<x-\cos^{-1}\dfrac35<0,$ $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=-\left(x-\cos^{-1}\dfrac35\right)$

For $-2\pi<x-\cos^{-1}\dfrac35<-\pi,$ $\cos^{-1}\bigg[\cos\Big(x-\cos^{-1}\dfrac35\Big)\bigg]=2\pi+x-\cos^{-1}\dfrac35$