Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    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!
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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?
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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 :/
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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?
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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}
    • Thread Starter
    Offline

    3
    ReputationRep:
    (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!
    • Community Assistant
    • Study Helper
    Offline

    20
    ReputationRep:
    Community Assistant
    Study Helper
    (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
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.