3sinx =2cosx ro find x.To find x.

Solution:

3sinx-2cosx = 0. Or

[3/sqrt(3^2+4^2)]sinx - [2/sqrt(3^2+2^2)]cosx = 0 . Or

sinxcosy-cosx siny = 0. where = arc tan 2/3 and siny =2/sqr13 and siny = 3/sqrt13.

sin(x-y) = 0. Or

x-y = npi. Or

x = y+npi = arc (tan2/3)+npi = 33.69006753 + n(180) degree.

We could solve the problem in 2 ways:

First method:

3 sin x = 2 cos x

sin x =( 2/3) cos x

sinx/cosx=2/3

tg x= 2/3

This is an elementary equation and we'll get:

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

The second method:

We apply the fundamental formula of trigonometry

sin^2 x + cos^2 x = 1

sin x = sqrt[(1 - (cosx)^2]

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

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

[(2/3)cos (x)]^2 = {sqrt[(1 - (cosx)^2]}^2

(4/9)(cosx)^2= 1 -(cosx)^2

(4/9)(cosx)^2+(cosx)^2 = 1

It is obvious that the same denominator is 9, so we'll multiply by 9, (cosx)^2, and we'll get:

(13/9)(cosx)^2= 1

(cosx)^2 = 9/13

cos x = 3*sqrt13/13

Again we'll have an elementary equation:

**x = arccos (3*sqrt13/13) + 2*k*pi**