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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`


`(cosx - sinx)^2=0`

`cosx - sinx = 0`


(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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jeyaram | Student


we know that sinx=cos(pi/2-x)

so sin2x=cos2x



(pi/2-2x)=2npi +or- 2x     n=Z

because if cosA=cosB

A=2npi +or- B


so get (-)  (pi/2-2x)=2npi - 2x     n=Z

in this time x is canceled both side

then get (+)  (pi/2-2x)=2npi + 2x  n=Z

4x=pi/2-2npi  n=Z


n=(-4)   x=17pi/8 is controdiction  (0<x<2pi)

n=(-3)   x=13pi/8

n=(-2)    x=9pi/8

n=(-1)    x=5pi/8

n=0       x=pi/8

n=1       x=(-3pi/8) is controdiction  (0<x<2pi)


so Answer is

x=13pi/8  or  9pi/8  or  5pi/8  or  pi/8