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For the following problem(s) I cannot get any answer(s). I would appreciate your help very much.

$$\tan { \theta -\sec { \theta } =\sqrt { 3 } } $$

TI get 30 degrees as the reference angle. What am I doing wrong because the answer is 210 degrees.

Link for my work since the post is not showing up.

Thanks for the help.

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    $\begingroup$ Be sure that your answer makes sense with the quadrant where the angle is placed. $\endgroup$ – Ángel Mario Gallegos Aug 23 '15 at 21:19
  • $\begingroup$ Posting a picture instead of writing the problem is discouraged and likely to be downvoted, can you make your post self-contained? $\endgroup$ – Alessandro Codenotti Aug 23 '15 at 21:24
  • $\begingroup$ @MarioG I was thinking that, but usually I find the answer by solving it. Since it equals root3/3 shouldn't the answer be 30 degrees? $\endgroup$ – Asker123 Aug 23 '15 at 21:29
  • $\begingroup$ The function tangent is not one to one, you are right with $\tan x=\frac{\sqrt{3}}{3}$, but there are two angles not coterminals satisfaying the equation, namely: $30^{\circ}$ and $210^{\circ}$. $\endgroup$ – Ángel Mario Gallegos Aug 23 '15 at 21:41
  • $\begingroup$ @MarioG Yes but 30 does not work for some reason. $\endgroup$ – Asker123 Aug 23 '15 at 21:43
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$$\tan { \theta -\sec { \theta } =\sqrt { 3 } } \\ \frac { \sin { \theta } }{ \cos { \theta } } -\frac { 1 }{ \cos { \theta } } =\sqrt { 3 } \\ \sin { \theta -\sqrt { 3 } \cos { \theta =1 } } \\$$ divide both side to $2$ $$ \frac { 1 }{ 2 } \sin { \theta -\frac { \sqrt { 3 } }{ 2 } } \cos { \theta } =\frac { 1 }{ 2 } \\ \sin { \frac { \pi }{ 6 } \sin { \theta } -\cos { \frac { \pi }{ 6 } \cos { \theta =\frac { 1 }{ 2\\ } } } } \\ \cos { \left( \frac { \pi }{ 6 } +\theta \right) =-\frac { 1 }{ 2 } } \\ \frac { \pi }{ 6 } +\theta =\pm \arccos { \left( -\frac { 1 }{ 2 } \right)+2n\pi =\pm \left( \pi -\frac { \pi }{ 3 } \right) } +2n\pi \\ \theta =\pm \frac { 2\pi }{ 3 } -\frac { \pi }{ 6 } +2n\pi,n\in\Bbb Z\ $$

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  • $\begingroup$ I don't quite understand the last part. Any help would be appreciated. $\endgroup$ – Asker123 Aug 27 '15 at 1:44
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$$\tan x-\sec x=\sqrt3\iff\tan x+\sec x=-\dfrac1{\sqrt3}$$

Adding we get $\tan x=\dfrac1{\sqrt3}\implies x=n\pi+\dfrac\pi6\ \ \ \ (1)$

Subtracting we get $\sec x=-\dfrac2{\sqrt3}\iff\cos x=-\dfrac{\sqrt3}2=\cos\left(\pi-\dfrac\pi6\right)$

$\implies x=2m\pi\pm\left(\pi-\dfrac\pi6\right)$

i.e, $=(2m+1)\pi-\dfrac\pi6\ \ \ \ (2)$ or $=(2m-1)\pi+\dfrac\pi6\ \ \ \ (3)$

$(1),(2),(3)\implies x=(2r-1)\pi+\dfrac\pi6$

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HINT:

Using Weierstrass substitution $$\dfrac{2t}{1-t^2}-\dfrac{1+t^2}{1-t^2}=\sqrt3$$ where $t=\tan\dfrac\theta2$

Rearrange to form a Quadratic equation in $t$

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Hint:

the solutions of $\tan x= \dfrac{\sqrt{3}}{3}$ are $x=\dfrac{\pi}{6}+k\pi$ (or in degrees: $x=30°+k 180°$).

Since you have squared to find the final equation, we must test the solutions in the starting equation.

For $x= \dfrac{\pi}{6}$ we find : $$ \dfrac{\sqrt{3}}{3}-2\dfrac{\sqrt{3}}{3}=-\dfrac{\sqrt{3}}{3} \ne \sqrt{3} $$ so it is an improper solution.

For $x= \dfrac{\pi}{6}+\pi$ we find : $$ \dfrac{\sqrt{3}}{3}+2\dfrac{\sqrt{3}}{3}= \sqrt{3} $$ so this is the principal solution.

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  • $\begingroup$ I got that answer and noticed I did something wrong. However I can't seem to find it. $\endgroup$ – Asker123 Aug 23 '15 at 21:36
  • $\begingroup$ I have corrected my answer. I hope that now it is usefull. $\endgroup$ – Emilio Novati Aug 24 '15 at 9:47

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