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

1. 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 ????
Attached Images

2. 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.
3. Hint: Plus or minus square root
4. (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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