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

    8
    ReputationRep:
    (Original post by hassi94)
    That's because it's trying to do tan(90) (undefined) then doing 2/it, instead of just doing 2cos(90) then dividing by sin90.
    I see what you did there:adore:
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    210 is, but it seems you've used an incorrect method (just got one correct answer coincidentally).

    Right so if I ask you to write down the values of 3x/2 for the range 0<x<360 what are your answers?
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
    Offline

    2
    ReputationRep:
    tan(3x/2)=-1
    a=3x/2
    tan(a)=-1 => a=3(pi)/4 + k(pi), k is an integer.
    3x/2=3(pi)/4 + k(pi)
    3x=3(pi)/2 + 2k(pi)
    x=(pi)/2 +2k(pi)/3, k is an integer.
    Offline

    14
    ReputationRep:
    (Original post by GreenLantern1)
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Offline

    14
    ReputationRep:
    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Yep.

    Now do you still need help with that other question?
    Offline

    0
    ReputationRep:
    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Do you know the general solution for tangent?

    If you consider the graph of tanx, and x = 0.5,
    http://www.wolframalpha.com/input/?i=plot+tanx+and+0.5
    they meet once in every period.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    Yep.

    Now do you still need help with that other question?
    Yes please

    I am just clueless as to how to approach it!
    Offline

    14
    ReputationRep:
    (Original post by GreenLantern1)
    Yes please

    I am just clueless as to how to approach it!
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    So the angles in answer to your question are x= 0,180,360,66.4,293.6
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    I also did another question: 2sin^2x=3cosx and i got values of 60,300 after making it into a quadratic. Is that correct; I think it is?
    Offline

    14
    ReputationRep:
    (Original post by GreenLantern1)
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by hassi94)
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
    Thanks, I was just a bit stuck at first
 
 
 
  • 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
    Has a teacher ever helped you cheat?
    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

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