Question

\sqrt{3} tg(\frac{\pi}{3}-x)-1\leq 0

Answer 1

$\displaystyle \sqrt{3} tg(\frac{\pi}{3}-x)-1\leq 0\\\\tg( \frac{ \pi }{3}-x) \leq \frac{1}{ \sqrt{3}}\\\\ * - \frac{ \pi }{2} \ \textless \ \frac{ \pi }{3}-x \leq \frac{ \pi }{6}; (+2 \pi n; n\in Z)\\\\ **\frac{ \pi }{2} \ \textless \ \frac{ \pi }{3}-x \leq \frac{7 \pi }{6}; (+2 \pi n; n\in Z)\\\\ * - \frac{ \pi }{2}- \frac{ \pi }{3} \ \textless \ -x \leq \frac{ \pi }{6}- \frac{ \pi }{3}; (+2 \pi n; n\in Z)\\\\ ** \frac{ \pi }{2}- \frac{ \pi }{3} \ \textless \ -x \leq \frac{7 \pi }{6}- \frac{ \pi }{3}; (+2 \pi n; n\in Z)$

$\displaystyle * \frac{ \pi }{6}+2 \pi n \leq x \ \textless \ \frac{5 \pi }{6}+2 \pi n; n\in Z\\\\ ** - \frac{5 \pi }{6} \leq x \ \textless \ - \frac{ \pi }{6}+2 \pi n; n\in Z$

Answer 2

To solve the inequality $\sqrt{3} \tan\left(\frac{\pi}{3}-x\right) - 1 \leq 0$, you can follow these steps:

1. First, isolate the term involving the tangent function on one side of the inequality: $\sqrt{3} \tan\left(\frac{\pi}{3}-x\right) \leq 1$

2. Divide both sides by $\sqrt{3}$ to get rid of the coefficient: $\tan\left(\frac{\pi}{3}-x\right) \leq \frac{1}{\sqrt{3}}$

3. Take the inverse tangent (arctan) of both sides to solve for $x$: $\frac{\pi}{3} - x \leq \arctan\left(\frac{1}{\sqrt{3}}\right)$

4. Subtract $\frac{\pi}{3}$ from both sides to isolate $x$: $-x \leq \arctan\left(\frac{1}{\sqrt{3}}\right) - \frac{\pi}{3}$

5. Finally, multiply both sides by -1 to reverse the inequality (and remember to reverse the direction of the inequality sign): $x \geq \frac{\pi}{3} - \arctan\left(\frac{1}{\sqrt{3}}\right)$

Now you have the solution for $x$: $x \geq \frac{\pi}{3} - \arctan\left(\frac{1}{\sqrt{3}}\right)$

You can calculate the numerical value of $\frac{\pi}{3} - \arctan\left(\frac{1}{\sqrt{3}}\right)$ to get the final answer.