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Mathematics 20 Online
OpenStudy (anonymous):

Solve: sin(2x)=sinx

myininaya (myininaya):

x=2npi n is an integer

myininaya (myininaya):

2x=x

myininaya (myininaya):

2x-x=0 1x=0 x=0

OpenStudy (anonymous):

double angle formula for sin2x: sin2x=sinx 2sinxcosx=sinx sinx=0 or cosx=1/2 x=0,2pi,pi/3,5pi/3

myininaya (myininaya):

largrange includes other solutions so he is right

myininaya (myininaya):

we have 2sinxcosx=sinx 2sinxcosx-sinx=0 sinx(2cosx-1)=0

myininaya (myininaya):

sinx=0 or 2cosx-1=0

OpenStudy (anonymous):

myininaya is also right, her answer is very thorough

OpenStudy (anonymous):

\[\sin(2x)=2\sin(x)\cos(x)\] \[\sin(2x)=\sin(x) \iff 2\sin(x)\cos(x)=\sin(x) \iff 2\cos(x)=1 \iff \cos(x)=\frac{1}{2}\] \[\iff x=\frac{\pi}{3} + 2k \pi, k \in \mathbb{Z} \ or x=-\frac{\pi}{3}+2k \pi, k \in \mathbb{Z}\]

myininaya (myininaya):

x=2npi, n is an integer cosx=1/2 when x=pi/3+2npi and x=5pi/3+2npi

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