Can you solve the equation for this interval [0, 2pi]? Sin(2x-(pi/4)) = (sqrt2)/(2)

2 Answers
Apr 6, 2018

#x = pi/4, pi/2, (5pi)/4, (3pi)/2#

Here's how I did it:

Explanation:

#sin(2x-(pi/4))=sqrt2/2#

First, we can rewrite this with an #sin^-1#
#sin^-1(sqrt2/2) = 2x - pi/4#

From the unit circle, we know that #sin(pi/4) = sqrt2/2#, meaning that #sin^-1(sqrt2/2) = pi/4#:

#pi/4 = 2x - pi/4#

Add #pi/4# to both sides of the equation:
#(2pi)/4 = 2x#
#pi/2 = 2x#

#x = (pi/2)/2#

#x = pi/2 * 1/2#

#x = pi/4#

There is still another solution since sine is also positive in the second quadrant:
To find the other solution, we refer to the unit circle:
enter image source here
We can see that #sinx = sqrt2/2# also occurs at #(3pi)/4#. So:
#2x-pi/4 = (3pi)/4#

#2x = 4pi/4#

#2x = pi#

#x = pi/2#

So we have #x = pi/4# and #pi/2#.

However, we want the solutions between #[0, 2pi]#.

First, we have to find the period of the function #sin(2x-pi/4)#. Recall that the period of #sinx = 2pi#. In this case, we are finding the period of #sin2x#, so:
#2pi/2 = pi#

The period is #pi#.

This means that we can add #pi# to each of our solutions:
#pi/4 + pi = pi/4 + (4pi)/4 = (5pi)/4#

#pi/2 + pi = pi/2 + (2pi)/2 = (3pi)/2#

We cannot add another #pi# to any of these solutions or else it would go over #[0, 2pi]#

So the final answers are #x = pi/4, pi/2, (5pi)/4, (3pi)/2#.

Hope this helps!

Apr 6, 2018

#color(blue)(pi/4 , (3pi)/2 , pi/2 ,(5pi)/4 )#

Explanation:

#sin(2x-pi/4)=sqrt(2)/2#

#arcsin(sin(2x-pi/4))=arcsin(sqrt(2)/2)=pi/4+n2pi,(3pi)/4+n2pi#

#2x-pi/4=pi/4+n2pi#

#2x=pi/2+n2pi#

#x=pi/4+npi#

#2x-pi/4=(3pi)/4+n2pi#

#2x=pi+n2pi#

#x=pi/2+npi#

For:

#[0,2pi]#

#color(blue)(pi/4 , (3pi)/2 , pi/2 ,(5pi)/4 )#