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Where have I gone wrong? *solving trig equation* Watch

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    This is how I solved it:
    cos2x-sin2x/ cos2x + 2 = 0
    so I multiplied all of them by cos2x
    cos2x-sin2x + 2cos2x = 0
    -sin2x + 3cos2x = 0
    -sin2x + 3(1-sin2x) = 0
    -sin2x + 3 -3sin2x = 0
    3 -4sin2x = 0
    4sin2x = 3
    sin2x = 3/4
    sinx = √3/2
    x= sin-1(√3/2)
    x= π/3 , 2π/3




    BUT!!
    In the markscheme, there are 2 additional answers 4π/3 and 5π/3 ????
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    When you square root you also need to look for solutions of sin(x)=sqrt(3)/2 since when you square the terms this is also a solution in the interval.
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    Hint: Plus or minus square root
    • Thread Starter
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    (Original post by poorform)
    When you square root you also need to look for solutions of sin(x)=sqrt(3)/2 since when you square the terms this is also a solution in the interval.
    (Original post by AnIndianGuy)
    Hint: Plus or minus square root
    Ohh!! I see thaaank you!!
 
 
 
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