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

1 Answer
Feb 21, 2015

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]}