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

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)