The Student Room Group

C2 Trig Question

1) Solve the following equation for t (theta), giving solutions in the interval 0 deg =< t =< 360 deg.
a) sin 2t - root 3 cos 2t = 0.

I don't know how to start this, because I can't see a use for any trig identity here :s-smilie: Any ideas? :rolleyes:
Reply 1
By the time I go and grab a piece of paper and pen, someone will already have answered you :smile:
Reply 2
Try to move the -root3 cos 2t to the other side, and divide both sides by cost2t

So you have tan2t=root3
Reply 3
Divide everything by cos2t ... should sort you out!
Reply 4
Original post by cpdavis
Try to move the -root3 cos 2t to the other side, and divide both sides by cost2t

So you have tan2t=root3


But then you lose solutions :rolleyes:
Reply 5
Original post by bcrazy
Divide everything by cos2t ... should sort you out!


You then lose solutions =/
Reply 6
Original post by Faith01
By the time I go and grab a piece of paper and pen, someone will already have answered you :smile:


Haha almost :biggrin:
Reply 7
Original post by adil12
Haha almost :biggrin:


I agree with the 2 people who already commented above..you don't lose solutions.
you dont lose solutions as 0 is not a solution.
Reply 9
Original post by Faith01
I agree with the 2 people who already commented above..you don't lose solutions.


Ohhhh of course my bad! Thanks everyone :smile:
Reply 10
Another way to do this problem would be to write it in the form Rsin(2tα)R\sin (2t-\alpha), but it seems like overkill for this question.

Original post by adil12
Ohhhh of course my bad! Thanks everyone :smile:


FWIW in case you want to know why you don't lose solutions: when you divide through by cos2t\cos 2t you're not cancelling a common factor, so it doesn't "disappear" as such, since it gets "caught" by the sin2t\sin 2t.
Reply 11
Original post by nuodai
Another way to do this problem would be to write it in the form Rsin(2tα)R\sin (2t-\alpha), but it seems like overkill for this question.



FWIW in case you want to know why you don't lose solutions: when you divide through by cos2t\cos 2t you're not cancelling a common factor, so it doesn't "disappear" as such, since it gets "caught" by the sin2t\sin 2t.


Yeah it seems too complex for this question xD

And I get what you mean, thank you for the additional info nuodai :smile:

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