3sinx = 2cosx

let us square both sides:

==> 9sin^2 x = 4cos^2 x..........(1)

We kow that :

sin^2 x + cos^2 x = 1

==> sin^2 x = 1 - cos^2 x

==> Substitute in (1):

==> 9(1-cos^2 x) = 4cos^2 x

==: 9 - 9cos^2 x = 4cos^2 x

Combine like terms:

==> 9- 13cos^2 x = 0

==> 13cos^2 x = 9

==: cos^2 x = 9/13

==> cosx = +-3/sqrt13

**==> x= arccos(3/sqrt13) + 2npi **

There are two methods of solving this issue:

First method:

3 sin x = 2 cos x

sin x =( 2/3) cos x

sinx/cosx=2/3

tg x= 2/3

x = arctg (2/3) + k*pi

The second method:

We know that in a right triangle, due to Pythagorean theorem,

sin^2 x + cos^2 x = 1

sin x = (1 - cos^2 (x))^1/2

But, from hypothesis, sin x = (2/3)cos x,so

(2/3)cos (x) = (1 - cos^2 (x))^1/2

[(2/3)cos (x)]^2 = [(1 - cos^2 (x))^1/2]^2

(4/9)cos^2 (x)= 1 - cos^2 (x)

(4/9)cos^2 (x )+ cos^2 (x) = 1

It is obvious that the same denominator is 9, so we'll multiply with 9, cos^2 (x) and the result will be:

(13/9)cos^2 (x) = 1

cos^2 (x) = 9/13

cos x = [3*(13)^1/2]/13

x = arccos {[3*(13)^1/2]/13} + 2*k*pi

3sinx = 2cosx

Divide by cosx.

sinx/cosx = 2/3

tanx = 2/3.

x = arctan (2/30

x = 33.69 deg. Or

x = 33.69 + 180 = 213.69 deg