sinx+cosx=cos2x,x∈[-π,π]求解三角方程
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sinx+cosx=√2 sin(x+π/4)
cos2x=sin(2x+π/2)=sin2(x+π/4)=2sin(x+π/4)*cos(x+π/4)
得√2 sin(x+π/4)= 2sin(x+π/4)*cos(x+π/4)
sin(x+π/4)=0 则x=-π/4或3π/4
sin(x+π/4)≠0 则cos(x+π/4)= √2/2 x=
所以方程的解为x= -π/2 -π/4 0 3π/4
cos2x=sin(2x+π/2)=sin2(x+π/4)=2sin(x+π/4)*cos(x+π/4)
得√2 sin(x+π/4)= 2sin(x+π/4)*cos(x+π/4)
sin(x+π/4)=0 则x=-π/4或3π/4
sin(x+π/4)≠0 则cos(x+π/4)= √2/2 x=
所以方程的解为x= -π/2 -π/4 0 3π/4
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