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

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We'll try to re-write the equation in a convenient way for solving:

sin 2x = 2sinx*cosx

cos 2x = (cos x)^2 - (sin x)^2

So, the equation will become:

2sinx*cosx = (cos x)^2 - (sin x)^2

We'll move the term 2sinx*cosx to the right side:

(cos x)^2 - 2sinx*cosx - (sin x)^2 = 0

We'll divide the equation by2sinx*cosx:

1 - 2sinx*cosx/2sinx*cosx - (sin x)^2 /(cos x)^2 = 0

We know that sinx/cosx=tg x

We'll substitute the ratio sin x/cos x by tg x:

1 - 2tgx - (tgx)^2=0

We'll multiply by (-1):

(tgx)^2 + 2tgx - 1 = 0

We'll note tgx=t

t^2 + 2t - 1 = 0

We'll apply the quadratic formula:

t1 = [-2+sqrt(4+4)]/2

t1 = 2(-1+sqrt2)/2

t1 = sqrt 2 - 1

t2 = -sqrt 2 - 1

tg x = t1

tg x = sqrt 2 - 1

**x = arctg (sqrt 2 - 1)**

**x = pi - arctg (sqrt 2 + 1)**

To solve sin2x= cos2x. x is in [0, pi]

Solution:

sin2x = cos2x = sin (Pi/2-2x).

2x=pi/2 -2x. Or

4x = pi/2. Or x = pi/8.

Similarly, sin2x = cos2x , when 2x= 225 or 5pi/4. Or

x = 5pi/8. Both solutions are in {0 , pi]

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