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x=5π6+kπx=\frac{5\pi}{6}+k\pi\newlineSolve the equation in the interval 0 \leq x < 2\pi.\newline3tanx3=03\tan x-\sqrt{3}=0

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Q. x=5π6+kπx=\frac{5\pi}{6}+k\pi\newlineSolve the equation in the interval 0x<2π0 \leq x < 2\pi.\newline3tanx3=03\tan x-\sqrt{3}=0
  1. Question Prompt: Question Prompt: Solve the equation 3tanx3=03\tan x - \sqrt{3} = 0 for xx in the interval 0 \leq x < 2\pi, given x=(5π6)+kπx = \left(\frac{5\pi}{6}\right) + k\pi.
  2. Step 11: Step 11: Simplify the given equation to find xx.3tanx=33\tan x = \sqrt{3}tanx=33\tan x = \frac{\sqrt{3}}{3}This implies x=arctan(33)x = \arctan(\frac{\sqrt{3}}{3}).

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