Question 1207278
<pre>
Ikleyn gave only one solution, but there are 4 solutions for {{{0<=x<360}}}
and infinitely many by adding 360<sup>o</sup>n to each of the four.
Every number is considered in degrees.

cos(2x+10)=sin(x+20)

This is a case of cos(A)=sin(B)

{{{cos(A)=sin(B)=cos(90-B)=cos(B-90)=cos(90-B+360n)=cos(B-90+360n)}}}

For the case 

{{{cos(A)=cos(90-B+360n)}}}
{{{A = 90-B+360n}}}
{{{2x+10=90-(x+20)+360n}}}
which simplifies to

{{{x=20+120n}}}

This gives us the solutions 20<sup>o</sup>, 140<sup>o</sup>, 260<sup>o</sup>,
between 0<sup>o</sup> and 360<sup>o</sup>.

However, this may not be all possible solutions {{{0^o<=x<=360^o}}}

We must also consider the case

{{{cos(A)=cos(B-90+360n)}}}
{{{A=B-90+360n}}}
{{{2x+10=(x+20)-90+360n}}}
which simplifies to
{{{x=360n-80}}}
This gives us only one additional solution 280<sup>o</sup> between 0<sup>o</sup> and 360<sup>o</sup>.

So all the solutions in {{{0<=x<360}}} are

20<sup>o</sup>, 140<sup>o</sup>, 260<sup>o</sup>, and 280<sup>o</sup>.

To get all infinitely many solutions, add 360n<sup>o</sup> to all four. 
(n is any integer, positive, negative or 0.)

Edwin</pre>