2sinx +cosx = 2.

We solve for x:

From the given equation we write 2sinx = 2-cosx

We square both sides:

4sin^2x = (4-4cosx +cos^2x)

4(1-cos^2x) = 4-4cosx+cos^2x

0 = -4cosx +5cos^2x

0 = cosx(-4+5cosx)

Therefore, cosx = 0 or -4+5cosx = 0.

cosx = 0 gives x = 2npi +pi/2 or 2npi-pi/2 , n = 0,1,2...

-4+5cosx = 0 gives: cosx = 4/5, or x = 360n +or- 36.86 deg nearly , when = 0,1, 2...

This is the linear equation and we'll solve it using the substitution technique.

We'll write 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]

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

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

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

We'll remove the brackets:

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

We'll move all terms to one side:

-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}**