Find all solutions on the interval [0, 2#pi#)? #2sin(2x)-1=0#

1 Answer
Apr 3, 2018

Answer:

#x=pi/12, x=(5pi)/12, x=(13pi)/12, x=(17pi)/12#

Explanation:

We start by adding 1 to both sides:

#2sin(2x)-cancel(1+1)=0+1#

#2sin(2x)=1#

Next we divide both sides by #2#:

#(cancel2sin(2x))/cancel2=1/2#

#sin(2x)=1/2#

We know that #sin(pi/6+2pik)=1/2#, which gives:

#2x=pi/6+2pik#

#(cancel2x)/cancel2=((pi/6+2pik)/2)#

#x=pi/12+pik#

This is one of the solutions. We can find the other by using the identity #sin(theta)=sin(pi-theta)#. This means that #sin(pi-pi/6+2pik)# also equals #1/2#, which gives the solution:

#2x=pi-pi/6+2pik#

#2x=(5pi)/6+2pik#

#(cancel2x)/cancel2=(5pi)/(2*6)+(2pik)/2#

#x=(5pi)/12+pik#

Now we need to find which of these solutions are inside the desired interval. We can write them out, seeing which ones are smaller than #2pi#:

#pi/12+pik->pi/12, (13pi)/12,(25pi)/12...#

#(5pi)/12+pik->(5pi)/12, (17pi)/12, (29pi)/12#

Everything larger than #(24pi)/12=2pi#, is not in the interval we want, so we can ignore those, leaving us with the following solutions:

#x=pi/12, x=(5pi)/12, x=(13pi)/12, x=(17pi)/12#