设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
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![设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已](/uploads/image/z/7606903-31-3.jpg?t=%E8%AE%BE0%E2%89%A4x%E2%89%A42%CF%80%2C%E4%B8%94%E2%88%9A%EF%BC%881-2sin+xcosx%3Dsin+x-cos+x%2C%E5%88%99+A.0%E2%89%A4x%E2%89%A4%CF%80+B%CF%80%2F4%E2%89%A4x%E2%89%A47%CF%80%2F4+C%CF%80%2F4%E2%89%A4x%E2%89%A45%CF%80%2F4+D%CF%80%2F2%E2%89%A4x%EF%BC%881%EF%BC%89%E8%AE%BE0%E2%89%A4x%E2%89%A42%CF%80%2C%E4%B8%94%E2%88%9A%EF%BC%881-2sin+xcosx%3Dsin+x-cos+x%2C%E5%88%99+A.0%E2%89%A4x%E2%89%A4%CF%80+B.%CF%80%2F4%E2%89%A4x%E2%89%A47%CF%80%2F4+C.%CF%80%2F4%E2%89%A4x%E2%89%A45%CF%80%2F4+D.%CF%80%2F2%E2%89%A4x%E2%89%A43%CF%80%2F2%EF%BC%882%EF%BC%89%E5%B7%B2)
设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x
(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则
A.0≤x≤π
B.π/4≤x≤7π/4
C.π/4≤x≤5π/4
D.π/2≤x≤3π/2
(2)已知sinθ+cosθ=1/5,θ∈(0,π)求sinθ,cosθ的值.
设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π Bπ/4≤x≤7π/4 Cπ/4≤x≤5π/4 Dπ/2≤x(1)设0≤x≤2π,且√(1-2sin xcosx=sin x-cos x,则 A.0≤x≤π B.π/4≤x≤7π/4 C.π/4≤x≤5π/4 D.π/2≤x≤3π/2(2)已
C
θ在第一象限
得sinθ>0,cosθ>0
θ在第二象限
得sinθ>0,cosθ
1,C,1-2sinxcosx=(sinx-cosx)(sinx-cosx)=sinx-cosx,因而有sinx=cosx或sinx-cosx=1;即x=π/4,5π/4;或x=π/2
2,sin=4/5,cos=-3/5
(1) 展开后应是sin(x/2)=cos(x/2)则x就该为π/4
(2) 就是把sinθ+cosθ=1/5平方后得到2sinθcosθ=-24/25 再用换元法就可以解出来了
(1)C