Angles of any magnitude- Question 3 (1 Viewer)

calebwhitaker199 · Jun 8, 2015

InteGrand · Jun 8, 2015

calebwhitaker199 said:

$$\sqrt{3}\tan^2 x = \tan x \Rightarrow \sqrt{3} \tan^2 x - \tan x = 0 \Rightarrow \tan x\left(\sqrt{3}\tan x - 1\right)=0$.$

$\therefore \tan x = 0$ or $\sqrt{3} \tan x - 1 = 0$

$Solutions for $\tan x = 0$ in the given domain are $x=-180^\circ,0^\circ, 180^\circ$.$

$Solutions for $\sqrt{3} \tan x - 1 = 0\Longleftrightarrow \tan x = \frac{1}{\sqrt{3}}$ in the given domain are $x=-150^\circ, 30^\circ$.$

$Therefore, the solutions are $x=-180^\circ, -150^\circ, 0^\circ, 30^\circ, 180^\circ$.$

leehuan · Jun 8, 2015

$\sqrt { 3 } \tan ^{ 2 }{ x } -\tan { x } =0\\ \tan { x } \left( \sqrt { 3 } \tan { x } -1 \right) =0\\ \tan { x } =0,\quad \sqrt { 3 } \tan { x } -1=0\quad \Rightarrow \tan { x } =\frac { 1 }{ \sqrt { 3 } }$

Pretty much done above, but the main thing is you must realise to not cancel out the tan(x) from both sides. Because in the process of that, you throw out tan(x)=0

calebwhitaker199 · Jun 8, 2015

Thanks guys

Angles of any magnitude- Question 3 (1 Viewer)

calebwhitaker199

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InteGrand

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leehuan

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calebwhitaker199

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