sin(x-π/3)=√2/2在[0,π]上的解是

已知函数f(x)=(√3sinωx+cosωx)*sin(-3π/2+ωx)（0 求证 sin(x+π/3)-√3cos(2π/3-x)+2sin(x-π/3)=0 已知函数f(x)=-√3sin^2x+sinxcosx 求在X∈[0,π/2]的值域 sin(x-π/3)=√2/2在[0,π]上的解是 f(x)=(√2 sin x + cos x)/(sin x +√(1- sin x)) (x∈ [0,π] 的最大值? 已知函数f（x）=［2sin（x-π/6）+√3sin x］cos x+sin^2x,x∈R f(x)=sin^2(wx)+√3sin(wx)*sin(wx+π/2）w>0的最小正周期为π,求w,f(x)在闭区间0,2π/3上取值范围 f(x)=sin^2(wx)+√3sin(wx)*sin(wx+π/2）w>0的最小正周期为π,求w,f(x)在闭区间0,2π/3上取值范围 已知-π/2＜x＜0,sin x+cos x=1/5求sin 2x+2 sin sin^2(π/3-x)+sin^2(π/6+x) 已知函数f(x)=sin²ωx+√(3)sinωx×cosωx（ω=2）,求f(x)在区间[0,2π/3]上的最大值和最小值 2sin(x+π/4)sin(x-π/4)=? 化简sin(x+7π/4)+cos(x-3π/4）步骤我已经找到撒sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2- 已知sin(x+π/6)=1/3,求sin(5π/6-x)+sin^2(π/3-x) 已知sin(x+π/6)=1/4,求sin(5π/6-x)+sin^2(π/3-x) sin平方x-sinxcosx-2cos平方x 求sin(x-π/3)X为锐角 sin平方x-sinxcosx-2cos平方x=0 求sin(x-π/3) 若tanx=2,则2sin平方x-3sin（π-x）sin（π/2-x）= 化简 sin²x+sin²(x+2π/3)+sin²(x-2π/3)