Question 17.2.12: Solve: √3cos x-sin x=1 ....
Solve: \sqrt{3} \cos x-\sin x=1 .
Step-by-Step
Given: \sqrt{3} \cos x-\sin x=1 .\qquad \qquad …(i)
Dividing both sides of (i) by \sqrt{(\sqrt{3})^{2}+(-1)^{2}} , i.e., by 2 , we get
\begin{aligned}& \frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x=\frac{1}{2} \\ \\\Rightarrow & \cos x \cos \frac{\pi}{6}-\sin x \sin \frac{\pi}{6}=\frac{1}{2} \\ \\\Rightarrow & \cos \left(x+\frac{\pi}{6}\right)=\cos \frac{\pi}{3} \left[\because \frac{1}{2}=\cos \frac{\pi}{3}\right]\\ \\\Rightarrow & \left(x+\frac{\pi}{6}\right)=2 n \pi \pm \frac{\pi}{3}, \text { where } n \in I [\because \cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha] \\ \\\Rightarrow & \left(x+\frac{\pi}{6}\right)=\left(2 n \pi+\frac{\pi}{3}\right) \text { or }\left(x+\frac{\pi}{6}\right)=\left(2 n \pi-\frac{\pi}{3}\right), \\ \\& x=\left(2 n \pi+\frac{\pi}{6}\right) \text { or } x=\left(2 n \pi-\frac{\pi}{2}\right), \text { where } n \in I .\end{aligned}
Hence, the general solution is x=\left(2 n \pi+\frac{\pi}{6}\right) or x=\left(2 n \pi-\frac{\pi}{2}\right) , where n \in I .
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