If 3sinx = 2cosx determine the value of x.

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