Monday, December 13, 2010

Solve for x : 4sinx - 3cosx = 0

We'll solve the problem using 2
methods


First method:


4 sin x
= 3 cos x


sin x =( 3/4) cos
x


We'll divide by cos
x:


sinx/cosx = 3/4


But the
ratio sinx/cosx = tan x


tan x=
3/4


x = arctan (3/4) +
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 = sqrt[1 - cos^2
(x)]


But, from hypothesis, sin x = (3/4)cos
x,so


(3/4)cos (x) = sqrt[1 - cos^2
(x)]


We'll square raise both
sides:


[(3/4)cos (x)]^2 = {sqrt[1 - cos^2
(x)]}^2


(9/16)cos^2 (x)= 1 - cos^2
(x)


(9/16)cos^2 (x )+ cos^2 (x) =
1


The least common denominator is 16, so we'll multiply
with 16, cos^2 (x) and the result will be:


(25/16)cos^2 (x)
= 1


cos^2 (x) = 16/25


cos x =
4/5


x = arccos (4/5) +
2*k*pi

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