Asked by Daemon
sin3x−sinx=0
Answers
Answered by
Steve
sin3x = sinx(2cos(2x)+1)
= sinx(2(1-2sin^2x)-1)
= sinx(1-4sin^2x)
= sinx - 4sin^3x
now you have sin^3(x) = 0
or, consider that sin(x)=0 at all multiples of pi.
Sin3x oscillates 3 times as fast, so it is also zero at multiples of pi.
= sinx(2(1-2sin^2x)-1)
= sinx(1-4sin^2x)
= sinx - 4sin^3x
now you have sin^3(x) = 0
or, consider that sin(x)=0 at all multiples of pi.
Sin3x oscillates 3 times as fast, so it is also zero at multiples of pi.
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