This is the linear equation in sin x and cos x and we'll solve it using the substitution technique.

We'll write the functions sin x and cos x with respect to tangent of the half-angle.

sin x = 2 tan(x/2)/[1+(tan x/2)^2]

cos x = [1-(tan x/2)^2]/[1+(tan x/2)^2]

To write easier the equation, we'll substitute tan (x/2) = t

sin x = 2 t/(1+t^2)

cos x = (1-t^2)/(1+t^2)

We'll re-write the equation in t:

2*2 t/(1+t^2) + (1-t^2)/(1+t^2) - 2 = 0

We'll multiply by (1+t^2), the right side term 2:

4 t + (1-t^2) - 2(1+t^2) = 0

We'll remove the brackets:

4t + 1 - t^2 - 2 - 2t^2 = 0

We'll combine like terms:

-3t^2 + 4t - 1 = 0

We'll multiply by -1:

3t^2 - 4t + 1 = 0

We'll apply the quadratic formula:

t1 = [4+sqrt(16-12)]/6

t1 = (4+2)/6

t1 = 1

t2 = (4-2)/6

t2 = 1/3

Now, we'll determine x:

tan (x/2) = 1

x/2 = arctan 1 + k*pi

x = 2arctan 1 + 2k*pi

x = 2*pi/4 + 2k*pi

x = pi/2 + 2k*pi

tan (x/2) = 1/3

x = 2arctan (1/3) + 2k*pi

**The solutions of the equation are:**

**{pi/2 + 2k*pi} U {2arctan (1/3) + 2k*pi}**

To so solve 2-cosx= 2sinx.

We know that sin^2x+cos^2 = 1 is an identity.

So cosx = sqrt(1-sin^2x).

Therefore 2- sqrt(1-sin^2x) = 2sinx.

-sqrt(1-s^2) = 2s-2, where s = sinx.

1-s^2 = 4s^2-8s+4.

subtract 1-s^2 from both sides:

0 = 4s^2-8s+4 +s^2 -1

0 = 5s^2-8s +3.

0 = (5s-3 )(3s -1)

5s - 3 = 0 , or 3s - 1 = 0

s = 3/5, or s = 1/3.

sinx = 3/5, or sinx = 1/3.

x = arcsin3/5, x = 36.87deg or x= 180-36.87 deg.

x = arcsin (1/3) , or x = 19.47 deg , or x = 1809 - 19.47 deg