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# (tan x)^2 − tan x − 42 = 0

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(tan x)^2 − tan x − 42 = 0

Guest Apr 17, 2015

#1
+18829
+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
Sort:

#1
+18829
+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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