# How do you solve #2cos x+3= sin2x# on the interval [0,2π]?

##### 1 Answer

It's impossible.

The functions sinus and cosine are functions that assume values in

The only *way* to have a solution is that it would be

so the possibilities are:

But the first member in

and in

So there are no solutions!

You can see also looking the graph of the function:

graph{2cosx+3-sin(2x) [-10, 10, -5, 5]}