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