solving trig equation

Solve for x:

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


Thanks.

Quote:

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Originally Posted by Mr_Green

Solve for x:

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


Thanks.

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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

to signify all answers would it be pi/6 + 2pi*K

thanks

Quote:

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Originally Posted by Mr_Green

to signify all answers would it be pi/6 + 2pi*K

thanks

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yea

Quote:

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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

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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