how to solve equation sinx-sin5x=0?

3 Answers | Add Yours

hala718's profile pic

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

Posted on

sinx-sin5x=0

We know that sinx-siny= 2cos(x+y)/2 * sin(x-y)/2

==> sinx-sin5x= 0

==> 2cos(3x) sin(-2x)=0

==> -2cos3x * sin2x =0

==> sin2x=0   or cos3x=0

when sin2x=0 ==> 2x= 2npi  n=0,1,2 ...

==> x= npi

when cos3x= 0 ==> 3x= (2n+1)pi   n=0,1,2,...

==> x= [(2n+1)/3]pi

Then x= {npi, (2n+1)pi}  n=0,1,2....

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the solution sinx -sin5x = 0. Or

sin5x = sinx = sin (x+2npi)..........(1)  Or

sin5x = sin {(2n+1)Pi-x}.............(2)

From (1) 5x = x+2npi Or

5x-x = 2npi Or

4x = 2npi. Or

x = npi/2, for n=0,1,2........

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

6x = (2n+1)Pi. Or

x= (2n+1)pi/6, for n =0,1,2....

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

It is a subtraction of 2 alike trigonometric functions and we'll transform it into a product.

2 cos [(x+5x)/2]*sin [(x-5x)/2]=0

2 cos 3x * sin(-2x) = 0

We'll divide by 2 and we'll put each factor from the product as being 0.

cos3x = 0

This is an elementary equation:

3x = +/-arccos 0 + k*pi

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

x=+/-pi/6 + k*pi, where k is an integer number.

We'll solve the second elementary equation:

 sin(-2x) = 0

-sin 2x = 0

x= (-1)^k*arcsin0 + k*pi

x=k*pi

The set of solutions:

S={k*pi}or{+/- pi/6 + k*pi}

We’ve answered 318,982 questions. We can answer yours, too.

Ask a question