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Basics?

If Sin 2t = 0

then sin-1 (0) = 2t

0 = 2t

t = 0 ?

But the markscheme says 2t = pi so t = pi/2... I must be missing something very basic here? I can't seem to spot it though?

Thanks
Reply 1
Original post by sabre2th1
If Sin 2t = 0

then sin-1 (0) = 2t

0 = 2t

t = 0 ?

But the markscheme says 2t = pi so t = pi/2... I must be missing something very basic here? I can't seem to spot it though?

Thanks


2t=sin1(0)    2t=,0,π,2π, 2t = \sin^{-1}(0) \implies 2t = \ldots, 0, \pi, 2 \pi, \ldots

It has infinity solutions, you need to look at the range given in the question.
Reply 2
Original post by sabre2th1
If Sin 2t = 0

then sin-1 (0) = 2t

0 = 2t

t = 0 ?

But the markscheme says 2t = pi so t = pi/2... I must be missing something very basic here? I can't seem to spot it though?

Thanks


sin(π)=sin(0)=0\sin(\pi) = \sin(0) = 0

Presumably the question asks for a value of t between 2 values, otherwise you have LOTS of answers you could give for t (infinitely many).
(edited 11 years ago)
Reply 3
Original post by raheem94
2t=sin1(0)    2t=,0,π,2π, 2t = \sin^{-1}(0) \implies 2t = \ldots, 0, \pi, 2 \pi, \ldots

It has infinity solutions, you need to look at the range given in the question.



Original post by Tla
sin(π)=sin(0)=0\sin(\pi) = \sin(0) = 0

Presumably the question asks for a value of t between 2 values, otherwise you have LOTS of answers you could give for t (infinitely many).


Damn :facepalm: I forgot basic C2.. Thanks !

btw raheem I would rep you but I can't

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