Find all solutions of the equation sin 2x = cos 2x, if x is in the interval [0,pi].

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sin 2x= cos 2x      x belongs to [0,pi]

==> (sin2x)^2=(cos2x)^2 ....(1)

We know that (sin2x)^2+(cos2x)^2=1

==> (sin2x)^2=1-(cos2x)^2

Now substitute in (1)

1-(cos2x)^2=(cos2x)^2

1=2(cos2x)^2

==> (cos2x)^2=1/2

==> cos2x= 1/sqrt2  

==> 2x= pi/2+2n*pi

==> x= pi/4+n*pi   n=0,1,2....

x= pi/4, pi/4+pi, pi/4+2*pi, .....

Since x belongs to the interval [0,pi],

==> x= pi/4...

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sin 2x= cos 2x      x belongs to [0,pi]

==> (sin2x)^2=(cos2x)^2 ....(1)

We know that (sin2x)^2+(cos2x)^2=1

==> (sin2x)^2=1-(cos2x)^2

Now substitute in (1)

1-(cos2x)^2=(cos2x)^2

1=2(cos2x)^2

==> (cos2x)^2=1/2

==> cos2x= 1/sqrt2  

==> 2x= pi/2+2n*pi

==> x= pi/4+n*pi   n=0,1,2....

x= pi/4, pi/4+pi, pi/4+2*pi, .....

Since x belongs to the interval [0,pi],

==> x= pi/4 =22.5 degrees

 

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