• How do you solve: sin2x = cos2x for 0 < or equal x < 2pi
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    Solve `sin2x=cos2x` for `0 <= x <2pi` .

    (1) Since `0<=x<2pi` we have `0<=2x<4pi` .

    (2) Use the identities `sin2x=2sinxcosx` and `cos2x=cos^2x-sin^2x`

    (3) Then `2sinxcosx=cos^2x-sin^2x`

    `cos^2x-2sinxcosx-sin^2x=0`

    `(cosx - sinx)^2=0`

    `cosx - sinx = 0`

    `cosx=sinx`

    (4) For the interval `[0,4pi)` `cosx=sinx` at `pi/4,(5pi)/4,(9pi)/4,(13pi)/4`

    (5) So on `[0,2pi)` `sin2x=cos2x` at `pi/8,(5pi)/8,(9pi)/8,(13pi)/8`

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