当sinx≠0时，令f(x)=sin(3x)/sinx = (sin2x*cosx+cos2x*sinx)/sinx
=(2sinx*cos²x+cos2x*sinx)/sinx
= 2cos²x + cos2x = 1+2cos2x
(1)当sin(π/4-α)≠0时，由原方程可知 f(π/4-α) = f(α)，则 cos(π/2-2α)=cos(2α)，即sin(2α) = cos(2α)
则2α = π/4+kπ，α = π/8+kπ/2，k∈Z
原方程还要求分母sinα≠0, sin(3α)≠0，检验成立
(2)当sin(π/4-α)=0 时要使原方程成立，则sinα≠0, sin(3α)≠0 且 sin(3π/4-3α)=0
所以存在k₁, k₂∈Z使 3π/4-3α =k₁π，π/4-α = k₂π，
解得α=π/4+kπ，k∈Z，此时sinα≠0, sin(3α)≠0，原方程成立
综上所述原方程的解是α=π/8+kπ/2 或者 π/4+kπ ，k∈Z
=(2sinx*cos²x+cos2x*sinx)/sinx
= 2cos²x + cos2x = 1+2cos2x
(1)当sin(π/4-α)≠0时，由原方程可知 f(π/4-α) = f(α)，则 cos(π/2-2α)=cos(2α)，即sin(2α) = cos(2α)
则2α = π/4+kπ，α = π/8+kπ/2，k∈Z
原方程还要求分母sinα≠0, sin(3α)≠0，检验成立
(2)当sin(π/4-α)=0 时要使原方程成立，则sinα≠0, sin(3α)≠0 且 sin(3π/4-3α)=0
所以存在k₁, k₂∈Z使 3π/4-3α =k₁π，π/4-α = k₂π，
解得α=π/4+kπ，k∈Z，此时sinα≠0, sin(3α)≠0，原方程成立
综上所述原方程的解是α=π/8+kπ/2 或者 π/4+kπ ，k∈Z
