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

1 Answer
Apr 3, 2018

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