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(tan x)^2 − tan x − 42 = 0

Guest Apr 17, 2015

Best Answer 

 #1
avatar+18712 
+10

(tan x)^2 − tan x − 42 = 0

We substitute:  $$\small{\text{$z=\tan{(x)}$}}$$

So we have: $$\small{\text{$z^2 -z - 42 = 0$}}$$

$$\small{\text{$
z_{1,2}=\frac{ 1\pm \sqrt{1-4\cdot(-42)} }{2\cdot 1}
=\frac{ 1\pm \sqrt{1+168 } }{2}
=\frac{ 1\pm \sqrt{169 } }{2}
=\frac{ 1\pm 13 }{2}
$}}\\\\
\small{\text{
$\begin{array}{l|l}
\hline
\\
z_1=\frac{ 1 + 13 }{2} \quad & \quad z_2=\frac{ 1 - 13 }{2}\\\\
z_1=\frac{ 14 }{2} \quad & \quad z_2=-\frac{ 12 }{2}\\\\
z_1=7 \quad & \quad z_2=-6\\\\
\hline
\\
\tan{(x_1)}=z_1=7 \quad & \quad \tan{(x_2)}=z_2=-6 \\\\
x_1 = \arctan(7) \quad & \quad x_2 = \arctan(-6)\\\\
\boxed{x_1 = 81.8698976458\ensurement{^{\circ}} \pm k\cdot 180\ensurement{^{\circ}}} \quad & \quad \boxed{ x_2 = -80.5376777920\ensurement{^{\circ}}\pm k\cdot 180\ensurement{^{\circ}}}
\end{array}
$}}$$
  

k= 0,1,2, ...

heureka  Apr 17, 2015
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1+0 Answers

 #1
avatar+18712 
+10
Best Answer

(tan x)^2 − tan x − 42 = 0

We substitute:  $$\small{\text{$z=\tan{(x)}$}}$$

So we have: $$\small{\text{$z^2 -z - 42 = 0$}}$$

$$\small{\text{$
z_{1,2}=\frac{ 1\pm \sqrt{1-4\cdot(-42)} }{2\cdot 1}
=\frac{ 1\pm \sqrt{1+168 } }{2}
=\frac{ 1\pm \sqrt{169 } }{2}
=\frac{ 1\pm 13 }{2}
$}}\\\\
\small{\text{
$\begin{array}{l|l}
\hline
\\
z_1=\frac{ 1 + 13 }{2} \quad & \quad z_2=\frac{ 1 - 13 }{2}\\\\
z_1=\frac{ 14 }{2} \quad & \quad z_2=-\frac{ 12 }{2}\\\\
z_1=7 \quad & \quad z_2=-6\\\\
\hline
\\
\tan{(x_1)}=z_1=7 \quad & \quad \tan{(x_2)}=z_2=-6 \\\\
x_1 = \arctan(7) \quad & \quad x_2 = \arctan(-6)\\\\
\boxed{x_1 = 81.8698976458\ensurement{^{\circ}} \pm k\cdot 180\ensurement{^{\circ}}} \quad & \quad \boxed{ x_2 = -80.5376777920\ensurement{^{\circ}}\pm k\cdot 180\ensurement{^{\circ}}}
\end{array}
$}}$$
  

k= 0,1,2, ...

heureka  Apr 17, 2015

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