Straw Hat Luffy
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Hey,

i was wondering if i could get some help on this question, I'm pretty lost and im not sure what identities would help most.

b) Solve, for 0 ≤ θ < 180, the equation

2cot²θ - 9cosecθ = 3

to 1 d.p.

Guidance would be very appreciated.

Thanks!
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Prince Philip
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(Original post by Straw Hat Luffy)
Hey,

i was wondering if i could get some help on this question, I'm pretty lost and im not sure what identities would help most.

b) Solve, for 0 ≤ θ < 180, the equation

2cot²θ - 9cosecθ = 3

to 1 d.p.

Guidance would be very appreciated.

Thanks!
Can you think of an identity involving \cot^2 \theta that you can use?
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Straw Hat Luffy
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(Original post by Notnek)
Can you think of an identity involving \cot^2 \theta that you can use?
I was thinking 1 + cot²θ = cosec²θ but I'm not sure how it'd be used in this, probably having a brainfart sorry :/
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Prince Philip
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(Original post by Straw Hat Luffy)
I was thinking 1 + cot²θ = cosec²θ but I'm not sure how it'd be used in this, probably having a brainfart sorry :/
Yes that's the one. You can rearrange that to give \cot^2 \theta = \mathrm{cosec}^2\theta - 1. Now subsitute this into your equation so you get

2(\mathrm{cosec}^2\theta - 1)-  9\mathrm{cosec} \ \theta  = 3

Now try expanding the brackets and simplifying. Post all your working if you get stuck.
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Straw Hat Luffy
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(Original post by Notnek)
Yes that's the one. You can rearrange that to give \cot^2 \theta = \mathrm{cosec}^2\theta - 1. Now subsitute this into your equation so you get

2(\mathrm{cosec}^2\theta - 1)-  9\mathrm{cosec} \ \theta  = 3

Now try expanding the brackets and simplifying. Post all your working if you get stuck.
Oh right cheers! I had a go but I feel i've gone wrong somewhere

I made the solution into a quadratic (2y²-9y-5=0) and got cosecθ values (I refered to as y) as 5 and - 1/2

I got sinθ= 1/5 and -2 respectively.

I attempt to do the inverse to θ however I get an error. Is it because i'm in the incorrect mode or have a made a stupid mistake somewhere.

Can you see if and where I've gone wrong?
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Prince Philip
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(Original post by Straw Hat Luffy)
Oh right cheers! I had a go but I feel i've gone wrong somewhere

I made the solution into a quadratic (2y²-9y-5=0) and got cosecθ values (I refered to as y) as 5 and - 1/2

I got sinθ= 1/5 and -2 respectively.

I attempt to do the inverse to θ however I get an error. Is it because i'm in the incorrect mode or have a made a stupid mistake somewhere.

Can you see if and where I've gone wrong?
You haven't gone wrong. \sin \theta = -2 has no solutions (you should know that by considering the sine graph) so this means you just ignore this equation. The only solutions will come from

\sin\theta = \frac{1}{5}
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Straw Hat Luffy
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(Original post by Notnek)
You haven't gone wrong. \sin \theta = -2 has no solutions (you should know that by considering the sine graph) so this means you just ignore this equation. The only solutions will come from

\sin\theta = \frac{1}{5}
Oh alright, forgot bout that! Did the inverse and got 11.53 degrees to 2 d.p. hopefully should be correct. You've been extremely helpful pal, thank you!
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Prince Philip
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(Original post by Straw Hat Luffy)
Oh alright, good to know! Did the inverse and got 11.53 degrees to 2 d.p. hopefully should be correct. You've been extremely helpful pal, thank you!
It's correct but you missed one solution.

No problem. Please continue to make threads whenever you need maths help
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