Question 971854
<pre>
{{{3 + cos(2x)}}}{{{""=""}}}{{{7/2}}}

Clear the fraction by multiplying every term by 2

{{{6 + 2cos(2x)}}}{{{""=""}}}{{{7}}}

Subtract 6 from both sides:

{{{2cos(2x)}}}{{{""=""}}}{{{1}}}

Divide both sides by 2

{{{cos(2x)}}}{{{""=""}}}{{{1/2}}}

Since {{{1/2}}} is positive, 2x is in QI or QIV

We want to include all but only all values of x on [0, 2&#960;)
which is the inequality:

{{{0}}}{{{""<=""}}}{{{x}}}{{{""<""}}}{{{2pi}}}

Since our equation involves 2x, not just x, we must multiply 
that inequality through by 2:

{{{0}}}{{{""<=""}}}{{{2x}}}{{{""<""}}}{{{4pi}}}

So all possibilities for 2x in that interval for

{{{cos(2x)}}}{{{""=""}}}{{{1/2}}}

are

{{{matrix(1,9,   2x,""="",pi/3,",",5pi/3,",",7pi/3,",",11pi/3)}}}

And we solve for x:

{{{matrix(1,9,   x,""="",pi/6,",",5pi/6,",",7pi/6,",",11pi/6)}}}

Edwn</pre>