# solving trig equation

• Mar 3rd 2007, 12:19 PM
Mr_Green
solving trig equation
Solve for x:

sin(2x) = sqrt(3) * sin(x)

Thanks.
• Mar 3rd 2007, 12:32 PM
Jhevon
Quote:

Originally Posted by Mr_Green
Solve for x:

sin(2x) = sqrt(3) * sin(x)

Thanks.

Use the trig identity sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

so sin(2x) = sqrt(3) * sin(x)
=> 2sinxcosx = sqrt(3)sinx
=> (2cosx)sinx = (sqrt(3))sinx .........i rewrote this so you would see the connection.

To maintain equality, 2cosx must be equal to sqrt(3), so the solution to our equation is the solution to 2cosx = sqrt(3)

=> cosx = sqrt(3)/2
=> x = pi/6
• Mar 3rd 2007, 12:47 PM
Mr_Green
to signify all answers would it be pi/6 + 2pi*K

thanks
• Mar 3rd 2007, 01:21 PM
Jhevon
Quote:

Originally Posted by Mr_Green
to signify all answers would it be pi/6 + 2pi*K

thanks

yea
• Mar 3rd 2007, 05:46 PM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
Use the trig identity sin(2x) = sinxcosx + sinxcosx = 2sinxcosx

so sin(2x) = sqrt(3) * sin(x)
=> 2sinxcosx = sqrt(3)sinx
=> (2cosx)sinx = (sqrt(3))sinx .........i rewrote this so you would see the connection.

To maintain equality, 2cosx must be equal to sqrt(3), so the solution to our equation is the solution to 2cosx = sqrt(3)

=> cosx = sqrt(3)/2
=> x = pi/6