Question

\[\sin(3x)+\cos(2x)-\sin(x)=0\]

Simplified sin(3x)
\[\sin(3x)=\sin(x+2x)=\sin x\cos(2x)+\cos x\sin(2x)=\sin x(\cos^2x-\sin^2x)+2\cos^2 x\sin x=\]
\[=\sin x(\cos^2x-sin^2x+2cos^2x)=\sin x(3\cos^2x-\sin^2x)\]

and then I get

\[\sin x(3\cos^2x-\sin^2x)+\cos^2x-\sin^2x-\sin x=0\]

and I don't know what else I may do.

Maybe there's some tricky thing I need to notice and then everything goes out?

If it's just really some very long and repetitive things needed to be done then don't worry replying because it's just "for fun".

Answer 1

WHERE'S EDIT BUTTON?
this is line which is not seen fully, if it matters
\[\sin(3x)=\sin(x+2x)=\sin x\cos(2x)+\cos x\sin(2x)=\]
\[=\sin x(\cos^2x-\sin^2x)+2\cos^2 x\sin x=\sin x(\cos^2x-sin^2x+2cos^2x)=\]
\[=\sin x(3\cos^2x-\sin^2x)\]

Answer 2

What if we use the sum formula:
sin(x)-sin(y)=2cos((x+y)/2)sin((x-y)/2)

So
sin(3x)-sin(x) + cos(2x)=0 =>
2cos(2x)sin(x)+cos(2x)=0 =>
cos(2x)[2sin(x)+1]=0 =>
cos(2x)=0 or sin(x)=-1/2

Answer 3

yeah, thats the trick lol :D

Answer 4

A Mathematica presentation is attached.