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C3 Trig Question: Where am I going wrong? watch

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    Question and my working out has attached

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    It should be 17cos(x + 1.08) = 17
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    (Original post by morgan8002)
    It should be 17cos(x + 1.08) = 17
    Oh okay

    So then I would so

    x + 1.08 = arccos(1)

    x= 0, pi, 2pi...

    And then what would I do?
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    (Original post by creativebuzz)
    Oh okay

    So then I would so

    x + 1.08 = arccos(1)

    x= 0, pi, 2pi...

    And then what would I do?
    x + 1.08 = 0, 2pi. (cos pi = -1)
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    (Original post by morgan8002)
    x + 1.08 = 0, 2pi. (cos pi = -1)
    Ah, I understand! thank you

    Is it alright if you could see what else I need to do in this question? Because I managed to get 3 of the answers but apparently I'm still missing 2 other answers (1.40 and 4.88)

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    (Original post by creativebuzz)
    Ah, I understand! thank you

    Is it alright if you could see what else I need to do in this question? Because I managed to get 3 of the answers but apparently I'm still missing 2 other answers (1.40 and 4.88)
    I would factorise rather than divide. You are at risk of losing solutions if you divide (like in this case).
    6\sin x\cos x = \sin x
    (\sin x)(6\cos x - 1) = 0
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    (Original post by morgan8002)
    I would factorise rather than divide. You are at risk of losing solutions if you divide (like in this case).
    6\sin x\cos x = \sin x
    (\sin x)(6\cos x - 1) = 0
    Ah okay, thanks! I assumed dividing was okay if it involved making tan, but I guess not! Thank you for your help
 
 
 
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