What are the all the solutions between 0 and 2π for #sin2x-1=0#?

2 Answers
Mar 7, 2016

Answer:

#x = pi/4# or #x = (5pi)/4#

Explanation:

#sin(2x) - 1 = 0#

#=> sin(2x) = 1#

#sin(theta) = 1# if and only if #theta = pi/2+2npi# for #n in ZZ#

#=>2x = pi/2+2npi#

#=> x = pi/4+npi#

Restricted to #[0, 2pi)# we have #n=0# or #n=1#, giving us

#x = pi/4# or #x = (5pi)/4#

Mar 7, 2016

Answer:

#S={pi/4,5pi/4}#

Explanation:

First, isolate the sine

#sin(2x)=1#

Now, take a look at your unit circle

etc.usf.edu

Now, the sine correspond to the #y# axis, so we can see that the only point between #0# and #2pi# where the sine is #1# is #pi/2# radians, so we have:

#2x = pi/2#

We want to solve for x, so

#x = pi/4#

However, remember that the period of the normal sine wave is #2pi#, but since we're working with #sin(2x)#, the period has changed; basically what we know is that there is a constant #k# that will act as the period, so:

#2(pi/4 + k) = pi/2 + 2pi#
#pi/2 + 2k = pi/2 + 2pi#
#2k = 2pi#
#k = pi#

And since #pi/4 + pi# or #5pi/4# is between #0# and #2pi#, that enters our set of solutions.

#S={pi/4,5pi/4}#