Find all solutions in the interval [0, 2π).

cos x = sin x

Hi, It is only to replace Cosx = Sinx

In the Identity equation :

(Cosx)^2 + (Sinx)^2 = 1

As Cos = Sin

Then,

(Sinx)^2 + (Sinx)^2 = 1

2.(Sinx)^2 = 1

Dividing both the sides by 2

(Sinx)^2 = 1/2

Applying square root on both the sides:

(Sinx) = √(1) / √(2)

Sinx = √(1)/√(2) × √(2)/√(2)

Sinx = √(2)/√(4)

Sinx = √(2)/2

Applying ArcSin on the sides of equation:

ArcSin(Sinx) = ArcSin( √(2)/2)

Canceling AcrSin with Sin

X = 45°

Or

As pi = 180°

Then applying the rule of 3

Pi 180°

Y 45°

Pi.45° = 180° .Y

180y = 45pi

y = 45pi/180

y = pi/4

As the Cos and Sin are igual in the 3 quadrant.

y = 45° + 180°

y = 225°

Or

y = pi/4 + pi

y = 5pi/4

Then,

y = pi/4 or 5pi/4

I hope this helped

r   = 2.65m

good luck!

cutie~