# C3 Trig Help (Maths)

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#1
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!
0
3 years ago
#2
(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 that you can use?
1
#3
(Original post by Notnek)
Can you think of an identity involving 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 :/
0
3 years ago
#4
(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 . Now subsitute this into your equation so you get

Now try expanding the brackets and simplifying. Post all your working if you get stuck.
0
#5
(Original post by Notnek)
Yes that's the one. You can rearrange that to give . Now subsitute this into your equation so you get

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?
0
3 years ago
#6
(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. 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

1
#7
(Original post by Notnek)
You haven't gone wrong. 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

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!
1
3 years ago
#8
(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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