Question

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

Answer 1

It's impossible.

The functions sinus and cosine are functions that assume values in `[-1,1]`, so the values of the first member are in `[1,4]`, but the second member still assumes values in `[-1,1]` (the `2x`, argument of the sinus, doesn't change the interval).

The only way to have a solution is that it would be `1`, the only one common value, but:

`sin2x=1rArr2x=pi/2+2kpirArrx=pi/4+kpi`

so the possibilities are: `pi/4,5/4pi`.

But the first member in `pi/4` assumes value:

`2cos(pi/4)+3=2sqrt2/2+3=sqrt2+3`, that isn't `1`,

and in `5/4pi`:

`2cos(5/4pi)+3=2(-sqrt2/2)+3=-sqrt2+3`, that isn't `1`.

So there are no solutions!

You can see also looking the graph of the function:

`y=2cosx+3-sin2x` and seeing that it doesn't touch the x-axis.

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