Question 389120
<pre>
You got two right and two wrong!

{{{3Cos(2x) + Cos(x) + 2 = 0}}}

Use the identity {{{Cos(2x)=2Cos^2x-1}}} 

{{{3(2Cos^2x-1) + Cos(x) + 2 = 0}}}

{{{6Cos^2x-3 + Cos(x) + 2 = 0}}}

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

Factor the left side:

{{{(3Cos(x)-1)(2Cos(x)+1)=0}}}

Use the zero-factor principle for the first factor:

{{{3Cos(x)-1=0}}}

{{{3Cos(x)=1}}}

{{{Cos(x) = 1/3}}}

We find the inverse cosine of {{{abs(1/3)}}} as 70.52877937°, which is
the reference angle.

Rounded to the nearest degree is 71°.  Since {{{1/3}}} is a positive
number and the Cosine is positive in the 1st and 4th quadrants, the
solutions are 71° for the 1st quadrant solution (same as the reference
angle) and 360°-70.52877937° = 289.4712206 or 289° for the 4th quadrant 
solution.


{{{(3Cos(x)-1)(2Cos(x)+1)=0}}}

Use the zero-factor principle for the second factor:

{{{2Cos(x)+1=0}}}

{{{2Cos(x)=-1}}}

{{{Cos(x) = -1/2}}}

We find the inverse cosine of {{{abs(-1/2)}}} as 60°, the reference
angle.

Since {{{-1/2}}} is a negative number and the Cosine is negative
in the 2nd and 3rd quadrants, the solutions are 180°-60° = 120° 
for the 2nd quadrant solution and 180°+60° = 240° for the 3rd quadrant
solution.

So the four correct answers to the nearest degree are:

71°, 289°, 120°, 240°

Edwin</pre>