Find the principal solution of the equation.
tan x = -√3
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We know that principal value branch of
![{tan}^{ - 1} x \:\: is\:\: [-\frac{\pi}{2}, \frac{\pi}{2}] \\](https://tex.z-dn.net/?f=+%7Btan%7D%5E%7B+-+1%7D+x+%5C%3A%5C%3A+is%5C%3A%5C%3A+%5B-%5Cfrac%7B%5Cpi%7D%7B2%7D%2C+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D+%5C%5C+ "{tan}^{ - 1} x \:\: is\:\: [-\frac{\pi}{2}, \frac{\pi}{2}] \\")
To solve the given equation we must keep consider that tan and tan inverse cancels each other only if x belongs to its principal value .
![tan \: x =- \sqrt{3} \\ \\ x = {tan}^{ - 1} (-\sqrt{3}) \\ \\ we \: know \: that \: tan \: \frac{\pi}{3} = \sqrt{3} \\ \\We\:\: know\:\:that\:\:\\\\tan \: ( - x) = - tan \: x \\ \\ {tan}^{ - 1} ( - tan \: x) = \pi -{tan}^{ - 1} ( tan \: x) \\\\ So\\\\ x= {tan}^{ - 1} (tan ( -\frac{\pi}{3} )) \\ \\x=\pi- {tan}^{ - 1} (tan ( \frac{\pi}{3} ))\\\\ x= \pi - \frac{ \pi}{3} \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2}, \frac{\pi}{2}]) \\ \\ x = \frac{2\pi}{3} \\](https://tex.z-dn.net/?f=tan+%5C%3A+x+%3D-+%5Csqrt%7B3%7D+%5C%5C+%5C%5C+x+%3D+%7Btan%7D%5E%7B+-+1%7D+%28-%5Csqrt%7B3%7D%29+%5C%5C+%5C%5C+we+%5C%3A+know+%5C%3A+that+%5C%3A+tan+%5C%3A+%5Cfrac%7B%5Cpi%7D%7B3%7D+%3D+%5Csqrt%7B3%7D+%5C%5C+%5C%5CWe%5C%3A%5C%3A+know%5C%3A%5C%3Athat%5C%3A%5C%3A%5C%5C%5C%5Ctan+%5C%3A+%28+-+x%29+%3D+-+tan+%5C%3A+x+%5C%5C+%5C%5C+%7Btan%7D%5E%7B+-+1%7D+%28+-+tan+%5C%3A+x%29+%3D+%5Cpi+-%7Btan%7D%5E%7B+-+1%7D+%28+tan+%5C%3A+x%29+%5C%5C%5C%5C+So%5C%5C%5C%5C+x%3D+%7Btan%7D%5E%7B+-+1%7D+%28tan+%28+-%5Cfrac%7B%5Cpi%7D%7B3%7D+%29%29+%5C%5C+%5C%5Cx%3D%5Cpi-+%7Btan%7D%5E%7B+-+1%7D+%28tan+%28+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29%29%5C%5C%5C%5C+x%3D+%5Cpi+-+%5Cfrac%7B+%5Cpi%7D%7B3%7D+%5C%3A+%5C%3A+%5C%3A+%5C%3A+%28x+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B-+%5Cfrac%7B%5Cpi%7D%7B2%7D%2C+%5Cfrac%7B%5Cpi%7D%7B2%7D%5D%29+%5C%5C+%5C%5C+x+%3D+%5Cfrac%7B2%5Cpi%7D%7B3%7D+%5C%5C+ "tan \: x =- \sqrt{3} \\ \\ x = {tan}^{ - 1} (-\sqrt{3}) \\ \\ we \: know \: that \: tan \: \frac{\pi}{3} = \sqrt{3} \\ \\We\:\: know\:\:that\:\:\\\\tan \: ( - x) = - tan \: x \\ \\ {tan}^{ - 1} ( - tan \: x) = \pi -{tan}^{ - 1} ( tan \: x) \\\\ So\\\\ x= {tan}^{ - 1} (tan ( -\frac{\pi}{3} )) \\ \\x=\pi- {tan}^{ - 1} (tan ( \frac{\pi}{3} ))\\\\ x= \pi - \frac{ \pi}{3} \: \: \: \: (x \: belongs \: to \: [- \frac{\pi}{2}, \frac{\pi}{2}]) \\ \\ x = \frac{2\pi}{3} \\")
is the solution of the equation.
To solve the given equation we must keep consider that tan and tan inverse cancels each other only if x belongs to its principal value .
is the solution of the equation.
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