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Math Help - General solution

  1. #1
    Junior Member
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    General solution

    Write general solutions:
    4sinacosa = 1
    the answer says (npi)/2 + (-1)^n (pi/12)
    except when i did it i got a different answer. if anybody could tell me where i went wrong i would be extremely grateful =D
    i rooted both sides : 2sinacosa = 1
    therefore, sin2a = 1
    2a = (pi n) + (-1) ^n (pi/2)
    divide by 2 both sides
    a = (pi n)/2 + (-1)^n (pi/4)

    Solve for 0 b 180
    b) secb + tan b = 2cosb
    i just realli dont no how to solve this
    answer is pi/6 and 5pi/6
    any help greatly appreciated ! thankyou !
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by THSKluv View Post
    4sinacosa = 1

    i rooted both sides : 2sinacosa = 1
    you got the wrong root
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  3. #3
    Member great_math's Avatar
    Joined
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    Quote Originally Posted by THSKluv View Post
    Write general solutions:
    4sinacosa = 1
    the answer says (npi)/2 + (-1)^n (pi/12)
    except when i did it i got a different answer. if anybody could tell me where i went wrong i would be extremely grateful =D
    i rooted both sides : 2sinacosa = 1
    therefore, sin2a = 1
    2a = (pi n) + (-1) ^n (pi/2)
    divide by 2 both sides
    a = (pi n)/2 + (-1)^n (pi/4)

    Solve for 0 b 180
    b) secb + tan b = 2cosb
    i just realli dont no how to solve this
    answer is pi/6 and 5pi/6
    any help greatly appreciated ! thankyou !
    4\sin a\cos a=1

    \Rightarrow 2\sin2a=1

    \Rightarrow \sin2a=\frac1{2}

    \Rightarrow \sin2a=\sin\frac\pi{6}

    \Rightarrow 2a=n\pi+(-1)^n\frac\pi{6}

    \Rightarrow a=\frac{n\pi}{2}+(-1)^n\frac\pi{12}
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  4. #4
    Junior Member
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    oh woops , yea that was stupid, sorry about that, please ignore my solution to the first part
    if anyone could help me solve it tho , reaaaly appreciate it thanks !
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  5. #5
    Junior Member
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    ahh ! thank you so much !
    i understand now =DD
    THANKS !
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