What is x if 2sinx + cosx = 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}