The given equation is  cos(2x+10) = sin(x+20).


It is assumed that the arguments of sine and cosine are in degrees.
Also, it is assumed, that the solution should be found in the interval 0 <= x < 360 degrees.


Use the identity cos(90-theta) = sin(theta)


Substituting the values we get,

    90-theta = 2x+10      (1)
       theta = x+20       (2)


We have simultaneous equations in two unknowns.
Exclude theta by adding equations 1 and 2

    90-theta+theta = 2x+10+x+20

    90 = 3x+30

    3x+30 = 90

    3x = 90-30 = 60

    x = 20


ANSWER.  x = 20 degrees.

Ikleyn gave only one solution, but there are 4 solutions for 
and infinitely many by adding 360on 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)



For the case 




which simplifies to



This gives us the solutions 20o, 140o, 260o,
between 0o and 360o.

However, this may not be all possible solutions 

We must also consider the case




which simplifies to

This gives us only one additional solution 280o between 0o and 360o.

So all the solutions in  are

20o, 140o, 260o, and 280o.

To get all infinitely many solutions, add 360no to all four. 
(n is any integer, positive, negative or 0.)

Edwin