If 3sinx = 2cosx determine the value of x.
5 Answers
hala718 | High School Teacher | (Level 1) Educator Emeritus
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
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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
madhavbhatia22 | Student, Grade 10 | (Level 1) eNoter
3sinx=2cosx
3/2=cosx/sinx
3/2tanx=0
3/2tan=1/x
=>x=2/3tan or x=3/2cot
madhavbhatia22 | Student, Grade 10 | (Level 1) eNoter
2/3tan
neela | High School Teacher | (Level 3) Valedictorian
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