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ind all solutions in the interval [0, 2π). 2 sin2x = sin x
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Double Angle \(\sin(2x) = \)??
Replace sin2x by 2sinxcosx
Gaaaa! ZeHanz gave it away!
There is still a lot of work left...
lol
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Okay then.
How did you replace it?
Or is it a fundemental identity?
It is rather fundamental I'm afraid...
Oh, okay, thanks.
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Just simplify by usoing fundemental identities?
What does it mean by the interval of (0, 2pi)?
Why not find ONE, first, before worrynig about that.
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