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3cos^2(x) - 7cos(x)+2 = 0

How do you solve?

 May 26, 2015

Best Answer 

 #1
avatar+33661 
+5

This factors as (3*cos(x) - 1)(cos(x) - 2) = 0

 

Since -1 ≤ cos(x) ≤ 1 the only valid solution is cos(x) = 1/3 so:

$${\mathtt{x}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{70.528\: \!779\: \!365\: \!509^{\circ}}}$$

 

This is a first quadrant solution. There is another solution in the fourth quadrant obtained by subtracting the above from 360°.

 May 26, 2015
 #1
avatar+33661 
+5
Best Answer

This factors as (3*cos(x) - 1)(cos(x) - 2) = 0

 

Since -1 ≤ cos(x) ≤ 1 the only valid solution is cos(x) = 1/3 so:

$${\mathtt{x}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{x}} = {\mathtt{70.528\: \!779\: \!365\: \!509^{\circ}}}$$

 

This is a first quadrant solution. There is another solution in the fourth quadrant obtained by subtracting the above from 360°.

Alan May 26, 2015

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