Question 814143
<pre>
sin(&#952; + 4°) = cos(&#952; + 6°)

Use the cofunction identity  cos(<font face="symbol">a</font>) = sin{90°-<font face="symbol">a</font>)

sin(&#952; + 4°) = sin[90° - (&#952; + 6°)]

sin(&#952; + 4°) = sin[90° - &#952; - 6°)]

sin(&#952; + 4°) = sin(84° - &#952;)

Then we use the fact that

if sin(&#945;) = sin(&#946;)

then one or both of these
hold

&#945; = &#946; + 360°n
&#945; = (180°-&#946;) + 360°n

for integer n

Using the first one
     &#945; = &#946; + 360°n
&#952; + 4° = 84° - &#952; + 360°n
    2&#952; = 80° + 360°n
     &#952; = 40° + 180°n 

letting n=0 give &#952; = 40°
letting n=1 give &#952; = 220°

Using the second one
     &#945; = 180° - &#946; + 360°n
&#952; + 4° = 180° - (84 - &#952;) + 360°n
&#952; + 4° = 180° - 84° + &#952; + 360°n
    4° = 96° + 360°n
  -92° = 360°n
  {{{-92/360}}} = n
   {{{-23/90}}} = n

That is extraneous because n is an integer.

So the only solutions between 0° and 360° are

&#952; = 40° and &#952; = 220°

and the general solution is  

&#952; = 40° + 180°n, for any integer n.

Edwin</pre>