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The most general solutions of 2sinx+2cosx=2(1-1/√2)

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2sinx+2cosx=2*2(-1/√2) 

=>2(sinx+1/√2)+2(cosx+1/√2)=2=20+20

=>sinx+1/√2=0 and cosx+1/√2=0

So sinx=sin(-π/4)

=>x = nπ - (-1)n​​​​​(π/4), where n€Z

And for cosx =-1/√2=cos(3π/4)

=>x=2kπ+(3π/4) and 2kπ-(3π/4)

Where k € Z

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