How do you solve #sin(2x)=cos(x)#?

1 Answer
Apr 28, 2016

#x=-\pi/2,\pi/6,\pi/2, or {5\pi}/6# plus multiples of #2\pi#.

Explanation:

I cannot tell whether you have learned the double-angle formulas yet so I will develop a solution without them.

First observe that #\cos u=\cos v# when #u+v=2n\pi# or #u-v=2n\pi# for any integer #n#.

We then have #\sin(2x)=\cos(\pi/2-2x)#. So:

#\cos(\pi/2-2x)=\cos(x)#.

Then we have the two possibilities defined above.

Possibility 1:

#u+v=(\pi/2-2x)+x=2n\pi#
#\pi/2-x=2n\pi#
#x=\pi/2-2n\pi#

So #x=\pi/2# plus multiples of #2\pi#

Possibility 2:

#u-v=(\pi/2-2x)-x=2n\pi#
#\pi/2-3x=2n\pi#
#x=\pi/6-{2n}/{3\pi}#

So #x=-\pi/2, \pi/6# or #{5\pi}/6# plus multiples of #2\pi#.

Put the two possibilities together to get the answer.