I'm trying to find the points t in (0,2[itex]\pi[/itex]) such that sint=sin4t. So I use the fact that sinA=sinB <==> A=B+2n[itex]\pi[/itex] ([itex]n\in\mathbb{Z}[/itex]), which yields t=2n[itex]\pi[/itex]/3 ([itex]n\in\mathbb{Z}[/itex]). The only solutions of this in (0,2[itex]\pi[/itex]) are 2[itex]\pi[/itex]/3 and 4[itex]\pi[/itex]/3.(adsbygoogle = window.adsbygoogle || []).push({});

However, there are 7 intersection points, says the "indirect method" of finding the zeros in (0,2[itex]\pi[/itex]) of sint-sin4t = sint+sin(-4t) =2sin(3t/2)cos(5t/2).

The number 7 is also confirmed by my calculator. So why doesn't the direct method using only the properties of sine, give the correct answer?!?!

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# Confusion with very basic algebra

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