sinx+cosx=cos2x,x∈[-π,π]求解三角方程
1个回答

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