Question 770001
<pre>
               9tan(2x) &#8722; 18cos(x) = 0

Divide through by 9

                 tan(2x) &#8722; 2cos(x) = 0

                    {{{sin(2x)/cos(2x)}}} - 2cos(x) = 0

                {{{(2sin(x)cos(x))/(1-2sin^2(x))}}} - 2cos(x) = 0

Divide through by 2

                 {{{(sin(x)cos(x))/(1-2sin^2(x))}}} - cos(x) = 0

Multiply through by LCD of 1 - 2sinē(x)

sin(x)cos(x) - cos(x)(1 - 2sinē(x)) = 0

Factor out cos(x)

     cos(x)[sin(x) - (1 - 2sinē(x)] = 0

      cos(x)[sin(x) - 1 + 2sinē(x)} = 0

Use the zero-factor property:

cos(x) = 0;            sin(x) - 1 + 2sinē(x) = 0

x = {{{pi/2}}}, {{{3pi/2}}};            2sinē(x) + sin(x) - 1 = 0     

                   [2sin(x) - 1][sin(x) + 1] = 0

               2sin(x) - 1 = 0            sin(x) + 1 = 0       
                   2sin(x) = 1                sin(x) = -1 
                    sin(x) = {{{1/2}}}                     x = {{{3pi/2}}}
                         x = {{{pi/6}}}, {{{5pi/6}}}  

There are 4 solutions in  [0, 2&#960;).

               {{{pi/6}}}, {{{pi/2}}}, {{{5pi/6}}}, and {{{3pi/2}}}                             

Edwin</pre>