Solve X for Sinx= Cos5x

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

sinx=cos5x

we know that sinx= cos(pi/2-x)

==> cos(pi/2 -x)=cos5x

==> cos(pi/2-x)-cos x=0

We know that cosx -cosy= 2sin(x+y)/2 sin(x-y)/2

==> 2 sin [(pi/2 -x + x)/2]*sin[(pi/2 - x - x)/2] = 0

==> 2sin (pi/4)*sin(pi/4 - x) = 0

==> sin(pi/4 - x) = 0   

pi/4 - x = n*pi   ,where n=0,1,2....

-x = -pi/4 + n*pi

x= pi/4 +npi

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To solve this equation, first we have to recall the fact that sine and cosine are complementary functions, so that:

cos x = sin (pi/2 - x) and sin x = cos (pi/2 - x) 

We'll substitute sin x by  cos (pi/2 - x), so that:

cos (pi/2 - x) = cos5x

cos (pi/2 - x) - cos5x = 0

We'll transform the difference into a product:

2 sin [(pi/2 - x + x)/2]sin[(pi/2 - x - x)/2] = 0

2sin (pi/4)sin(pi/4 - x) = 0

We'll divide by 2sin (pi/4) and we'll get:

sin(pi/4 - x) = 0

This is an elementary equation:

pi/4 - x = (-1)^k*arcsin0 + k*pi

-x = -pi/4 + k*pi

We'll multiply by -1:

x = pi/4 - k*pi

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To solve for sinx = cos 5x.

Solution:

sinx =cos5x = cos [ pi/2 -5x] . Or

sinx  =  sin[(2n+1)pi/2-5x] . Or

x = (2n+1)pi/2-5x. Or

6x = (2n+1)pi./2 Or

x = (2n+1)pi/12, for n = 0,1,2....

 

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