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Find all values of \(x\) such that \(0\leq x\leq 2\pi\) and \( \sin(x)-\cos(x) = \dfrac{1}{\sqrt{2}} .\)

 Oct 28, 2019
 #1
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sin (x) - cos(x)  = 1/√2         square both sides

 

sin^2(x) - 2sin(x)cos(x) + cos^2x   = 1/2

 

-2sin(x)cos(x) + 1  =  1/2

 

-2sin(x)cos(x)  =  -1/2

 

2sin(x)cos(x)  =  1/2

 

sin (2x)  = 1/2

 

sin (x)  =  1/2   at   pi/6   ,   5pi/6 ,  13pi/6  ,  17pi/6

 

So

 

sin (2x)  =  1/2    at    pi/12 , 5pi/12 , 13pi/12  and 17pi/12

 

As the graph here shows, the only solutions  occur at 5pi/12  and 13pi/12 : https://www.desmos.com/calculator/lknuil6pvy

 

[The other two solutions are extraneous because we squared the original equation ]

 

 

 

cool cool cool

 Oct 28, 2019

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