Find the principal and general solutions of the question $\tan x=\sqrt{3}$.
Find the principal and general solutions of the question $\tan x=\sqrt{3}$.
$\tan x=\sqrt{3}$
It is known that $\tan \frac{\pi}{3}=\sqrt{3}$ and $\tan \left(\frac{4 \pi}{3}\right)=\tan \left(\pi+\frac{\pi}{3}\right)=\tan \frac{\pi}{3}=\sqrt{3}$
Therefore, the principal solutions are $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$
Now, $\tan x=\tan \frac{\pi}{3}$
$\Rightarrow x=n \pi+\frac{\pi}{3},$ where $n \in Z$
Therefore, the general solution is $x=n \pi+\frac{\pi}{3},$ where $n \in Z.$
Similar Questions
Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy
$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$
$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$
Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy
$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$
$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$
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