Solve for x:

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

Thanks.

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- Mar 3rd 2007, 12:19 PMMr_Greensolving trig equation
Solve for x:

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

Thanks. - Mar 3rd 2007, 12:32 PMJhevon
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 PMMr_Green
to signify all answers would it be pi/6 + 2pi*K

thanks - Mar 3rd 2007, 01:21 PMJhevon
- Mar 3rd 2007, 05:46 PMThePerfectHacker
You made an error :eek: what about x=0?

Instead when you have,

(2cosx)sinx = (sqrt(3))sinx

You cannot divide by "sin x" which is what you implicity did.

Rather subtract to get,

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

Factor,

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

Thus,

sin x=0

2cos x-sqrt(3)=0