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Find all solutions in the interval [0, 2π). sec2 x - 2 = tan2 x I'm not sure how to do this, could someone walk me through?
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\[sec^2 x-2=tan^2 x?\]
or sec (2x) -2 = tan (2x)?
the first one :)
no solution, friend. since 1+ tan^2 x = sec^2 x ---> 2 + tan^2 x = 1+ 1 + tan^2x = 1 + sec^2x = sec^2x leads to 0=1 ( therefore, no solution)
oh ok i see now, thank you!
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