Question

number of integral value (s) of m for the equation $\sin x-\sqrt{3}\cos x=\frac{4m-6}{4-m}$ has solution $x\in \left[0,\pi \right]$.

Answer 1

According to question.........

$\begin{array}{c}\sin x-\sqrt{3}\cos x=\frac{4m-6}{4-m}\\ \Rightarrow 2\left(\frac{1}{2}x-\frac{\sqrt{3}}{2}\cos x\right)=\frac{4m-6}{4-m}\\ \Rightarrow 2\left(\cos \frac{\pi }{3}.\sin x-\sin \frac{\pi }{3}.cosx\right)=\frac{4m-6}{4-m}\\ \Rightarrow 2\left(\sin (x-\frac{\pi }{3})\right)=\frac{4m-6}{4-m}\\ \Rightarrow sin(x-\frac{\pi }{3})=\frac{2m-3}{4-m}|\begin{array}{c}-1\le sin(x-\frac{\pi }{3})\le 1\\ -1\le \frac{2m-3}{4-m}\le 1\end{array}\\ \frac{2m-3}{4-m}+1\ge 0---------\left(i\right)\\ \Rightarrow \frac{2m-3+4-m}{4-m}\ge 0\\ \Rightarrow \frac{m+1}{4-m}\ge 0,m\in \left[-1,4\right]\\ Again,\\ \frac{2m-3}{4-m}-1\le 0--------\left(ii\right)\\ \Rightarrow \frac{2m-3-4+m}{4-m}\le 0\\ \Rightarrow \frac{3m-7}{4-m}\le 0m\in \left[-\infty ,\frac{7}{3}\right]\cup \left[4,\infty \right]\\ so,thatm\in \left[-1,\frac{7}{3}\right]\\ m=-1,0,1,2\end{array}$

and there are 4 integers.