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C3 Trig

2cot23θ=7cosec3θ−52cot^2 3\theta = 7 cosec 3\theta -5

I think this turns out to be a quadratic if you sub in the cosec identity but I'm not sure.
Reply 1
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Notnek
21
Original post by zed963
2cot23θ=7cosec3θ−52cot^2 3\theta = 7 cosec 3\theta -5

I think this turns out to be a quadratic if you sub in the cosec identity but I'm not sure.

Yes it will - have you tried?

Can you have a go and post all your working if you get stuck.
Reply 2
Avatar for zed963
zed963
OP
16
Original post by notnek
Yes it will - have you tried?

Can you have a go and post all your working if you get stuck.


so far let's replace theta with x, it's easier to type.

2(cosec^2 3x-1)=7cosec 3x-5
2cosec^2 3x - 7cosec 3x +3 =0

From here factorise and solve.

Change the interval to 540 degrees.
Reply 3
Avatar for Notnek
Notnek
21
Original post by zed963
so far let's replace theta with x, it's easier to type.

2(cosec^2 3x-1)=7cosec 3x-5
2cosec^2 3x - 7cosec 3x +3 =0

From here factorise and solve.

Change the interval to 540 degrees.

That's correct so far. Are you able to continue from here?
Reply 4
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zed963
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Original post by notnek
That's correct so far. Are you able to continue from here?


Yeah when factorised

1/3=sin x (x=19.5)
1/0.5 = sin x ( no solutions )

so 3x = 19.5, 160.5, 379.5, 520.5
x = 6.5, 53.5, 126.5, 173.5

Is that correct?
Reply 5
Avatar for Notnek
Notnek
21
Original post by zed963
Yeah when factorised

1/3=sin x (x=19.5)
1/0.5 = sin x ( no solutions )

so 3x = 19.5, 160.5, 379.5, 520.5
x = 6.5, 53.5, 126.5, 173.5

Is that correct?

That's correct if the interval given in the question is 0<x<180.
Reply 6
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zed963
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16
Original post by notnek
That's correct if the interval given in the question is 0<x<180.


Yes that's the correct interval :smile:

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