Question 1161422
<pre>
{{{matrix(1,3,

12cos^2(x)-4cos(x)=1,",",0<=x<2pi)}}}

{{{12cos^2(x)-4cos(x)-1=0}}}

Factor the left side:

{{{(6cos(x)+1^"")(2cos(x)-1^"")=0}}}

6cos(x)+1=0;      2cos(x)-1=0
  6cos(x)=-1;       2cos(x)=1
   cos(x)=-1/6;      cos(x)=1/2

For cos(x)=-1/6 we know that x is in QII or QIII. First we find the angle in
QI that has +1/6 for its cosine. [It will not be a solution but we can get 
the others from that reference angle.

We find the inverse cosine of +1/6 to be 1.403348238.  That's the reference
angle.  To get the QII answer we subtract from &pi; 

&pi;-1.403348238 = 1.738244406, rounds to 1.74 <--QII answer

To get the QIII answer we add to &pi; 

&pi;+1.403348238 = 4.544940902, rounds to 4.54 <--QIII answer

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For cos(x)=1/2 we know that x is in QI or QIV.  Those are special angles
that we get off the unit circle.

They are π/3 and 5π/3

So there are 4 answers, two of them exact, and the other two approximated.

Edwin</pre>