How do you find all solutions of the equation in the interval $[0, 2 π)$ $[0,2pi)$ given $\sqrt{3} tan 3 x = 0$ $sqrt3tan3x=0$ ?

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Answer 1 · 2017-04-09T06:55:25

$x = {0, \frac{π}{3}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{5 π}{3}}$ $x={0,pi/3,(2pi)/3,pi,(4pi)/3,(5pi)/3}$

$\sqrt{2} tan 3 x = 0 \Rightarrow tan 3 x = 0$ $sqrt2tan3x=0=>tan3x=0$ as $\sqrt{3} \neq 0$ $sqrt3!=0$

Hence $3 x = n π$ $3x=npi$ or $x = \frac{n π}{3}$ $x=(npi)/3$

and in the interval $[0, 2 π)$ $[0,2pi)$

$x = {0, \frac{π}{3}, \frac{2 π}{3}, π, \frac{4 π}{3}, \frac{5 π}{3}}$ $x={0,pi/3,(2pi)/3,pi,(4pi)/3,(5pi)/3}$

How do you find all solutions of the equation in the interval [0,2pi)[0,2π)[0,2pi) given sqrt3tan3x=0√3tan3x=0sqrt3tan3x=0?