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[tex]\displaystyle\\ -3\tan^2x + 1 = 0\\\\ -3\tan^2x = -1\\\\ \tan^2x = \frac{-1}{-3} \\\\ \tan^2x = \frac{1}{3} ~~~~~ \Big|~\sqrt{~~~} \\\\ \sqrt{\tan^2x} = \sqrt{\frac{1}{3}} \\\\ \tan x = \Big| \frac{1}{ \sqrt{3} } \Big|\\\\ \tan x = \Big| \frac{\sqrt{3}}{ 3 } \Big|\\\\ \tan x = \frac{\sqrt{3}}{ 3 }~~~\Longrightarrow~~x_1= \frac{\pi}{6} + k\pi,~~k \in N \\\\ \tan x = -\frac{\sqrt{3}}{ 3 }~~~\Longrightarrow~~x_2= \pi-\frac{\pi}{6} + k\pi = \frac{5\pi}{6} + k\pi,~~k \in N \\\\ [/tex]



JeanaShupp

Answer: [tex]x=\frac{\pi}{6}+n\pi,n\ \epsilon\ N[/tex] and

[tex]\ x=\frac{5\pi}{6}+n\pi,n\ \epsilon\ N[/tex]


Step-by-step explanation:

The given equation is [tex]-3\tan^2(x)+1=0[/tex]

Subtract 1 on both sides, we get

[tex]-3\tan^2(x)=-1[/tex]

Divide -3 on both sides, we get

[tex]\tan^2(x)=\frac{1}{3}[/tex]

Taking square root on both sides, we get

[tex]\tan\ x=\pm\frac{1}{\sqrt{3}}[/tex]

We know that [tex]\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}[/tex]

Thus, when [tex]\tan\ x=\frac{1}{3}\ then\ x=\frac{\pi}{6}+n\pi,n\ \epsilon\ N[/tex]

When  [tex]\tan\ x=-\frac{1}{3}\ then\ x=\pi-\frac{\pi}{6}+n\pi=\frac{5\pi}{6}+n\pi,n\ \epsilon\ N[/tex]

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