Question 1187250
<pre>

{{{cos(3x)}}}{{{""=""}}}{{{cos(0.5x)}}}

Write both angles in terms of their average.

Their average is {{{(3x+0.5x)/2}}}{{{""=""}}}{{{3.5x/2}}}{{{""=""}}}{{{1.75x}}}

{{{3x = 1.75x+1.25x}}} and {{{0.5x = 1.75x-1.25x}}}

Substituting in original equation,

{{{cos(1.75x+1.25x)}}}{{{""=""}}}{{{cos(1.75x-1.25x)}}}

{{{cos(1.75x)cos(1.25x)-sin(1.75x)sin(1.25x))}}}{{{""=""}}}{{{cos(1.75x)cos(1.25x)+sin(1.75x)sin(1.25x))}}}

{{{cross(cos(1.75x)cos(1.25x))-sin(1.75x)sin(1.25x))}}}{{{""=""}}}{{{cross(cos(1.75x)cos(1.25x))+sin(1.75x)sin(1.25x))}}}

{{{-2sin(1.75x)sin(1.25x))}}}{{{""=""}}}{{{0}}}

{{{sin(1.75x)sin(1.25x))}}}{{{""=""}}}{{{0}}}

{{{sin(1.75x)}}}{{{""=""}}}{{{0}}}; {{{sin(1.25x))}}}{{{""=""}}}{{{0}}}

{{{1.75x}}}{{{""=""}}}{{{180^o*n}}};   {{{1.25x}}}{{{""=""}}}{{{180^o*n}}};

{{{1.75=1&3/4=7/4}}} and {{{1.25=1&1/4=5/4}}} so the above is

{{{expr(7/4)x}}}{{{""=""}}}{{{180^o*n}}};   {{{expr(5/4)x}}}{{{""=""}}}{{{180^o*n}}}

{{{x}}}{{{""=""}}}{{{expr(4/7)180^o*n}}};   {{{x}}}{{{""=""}}}{{{expr(4/5)180^o*n}}}

{{{x}}}{{{""=""}}}{{{(720/7)^o*n}}};   {{{x}}}{{{""=""}}}{{{144^o*n}}}

But since {{{0 <= x <= 180^o}}},

{{{0 <= (720/7)^o*n <= 180^o}}},    {{{0 <= 144^o*n <= 180^o}}}, 

{{{0 <= (720)^o*n <= 1260^o}}},    {{{0 <= n <= 180^o/144^o}}}, 

{{{0 <= n <= 1260^o/720^o}}},    {{{0 <= n <= 180^o/144^o}}}, 

{{{0 <= n <= 1.75}}},    {{{0 <= n <= 1.25}}}, 

Since n is an integer, n is either 0 or 1.

So the solutions are

{{{x}}}{{{""=""}}}{{{(720/7)^o*0}}};   {{{x}}}{{{""=""}}}{{{144^o*0}}}

that is, x=0, and

{{{x}}}{{{""=""}}}{{{(720/7)^o*1}}};   {{{x}}}{{{""=""}}}{{{144^o*1}}}

{{{x}}}{{{""=""}}}{{{(720/7)^o}}};   {{{x}}}{{{""=""}}}{{{144^o}}}

So there are 3 solutions.

{{{x=0^o}}}, {{{x=(720/7)^o}}}, and {{{x=144^o}}}

 

Edwin</pre>