Why Does the Equation Sin x = Sin 2x Also Yield the Solutions 0 and Pi?

[kuahji]

The question is "Solve each equation for exact solutions over the interval of [0,2pi).

Equation: sin x = sin 2x

I transformed the equation to

sin x = 2sin x cos x

then

sin x/2sin x = cos x

cos x = 1/2

Answer: pi/3, 5pi/3

but the book also has the answers 0, pi.

What must I do to find the other two solutions?

 

[arildno]

You ASSUMED that sin(x) was non-zero when you divided with it, since you can't divide by zero.

Since we see that the equation is ALSO fulfilled in the case sin(x)=0
(The equation reduces then to 0=2*0*cos(x) which is always true),

then the solutions of sin(x)=0 are also solutions of the original equation.
Guess what those numbers might be..

 

[kuahji]

Thanks for the explanation. :)

*edit* Figured it out now.