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4sinx-3cosx=0Find x 

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greynose | Student, Undergraduate | Honors

Posted June 4, 2011 at 12:50 AM via web

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4sinx-3cosx=0

Find x 

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giorgiana1976 | College Teacher | Valedictorian

Posted June 4, 2011 at 4:11 AM (Answer #2)

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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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