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could you please solve for x this equation? sinx - cosx = 1`/sqrt(2)`

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blehens | eNoter

Posted August 16, 2013 at 11:26 PM via web

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could you please solve for x this equation?

sinx - cosx = 1`/sqrt(2)`

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aruv | High School Teacher | Valedictorian

Posted August 17, 2013 at 2:30 AM (Answer #1)

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`sin(x)-cos(x)=1/sqrt(2)`

`(1/sqrt(2))sin(x)-(1/sqrt(2))cos(x)=(1/sqrt(2))(1/sqrt(2))`

`sin(x)cos(pi/4)-cos(x)sin(pi/4)=1/2`

`because `

`cos(pi/4)=sin(pi/4)=1/sqrt(2)`

`therefore`

`sin(x-pi/4)=sin(pi/6)`

`because`

`sinacosb-cosasinb=sin(a-b)`

`Thus`

`x-pi/4=npi+(-1)^n(pi/6)`

`x=npi+pi/4+(-1)^n(pi/6)`  and n is an integer.

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 17, 2013 at 2:52 AM (Answer #2)

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Solve `sinx-cosx=1/sqrt(2)` :

Square both sides (be aware that we might introduce extraneous solutions.):

`sin^2x-2sinxcosx+cos^2x=1/2` Use the Pythagorean relationship:

`1-2sinxcosx=1/2`

`sinxcosx=1/4`  

Use `sinucosv=1/2[sin(u+v)+sin(u-v)]`

`1/2[sin2x+sin0]=1/4`

`sin2x=1/2`

`2x=pi/6,(5pi)/6,(13pi)/6,(17pi)/6` so

`x=pi/12,(5pi)/12,(13pi)/12,(17pi)/12`

Checking we find that `x=pi/12,x=(17pi)/12` produce `-1/sqrt(2)` in the original equation so they are extraneous.

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The solutions are `x=(5pi)/12+2npi,(13pi)/12+2npi` for `n in ZZ` (n an integer)

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The graph:

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