已知方程2sinx π 3 2a-1
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原式=(sin²x+sinxcosx)/(2sin²x+cos²x)1=sin²x+cos²x=(tan²x+tanx)/(2tan
f(x)=ab=2sinx(1+sinx)+(cosx+sinx)(cosx-sinx)=2sinx+2sin^2x+cos^2x-sin^2x=2sinx+sin^2x+cos^2x=2sinx+1
当原方程有实数解时,a+3=(sinx-2)².∵-1≤sinx≤1.∴1≤(sinx-2)²≤9.∴1≤a+3≤9.∴-2≤a≤6.
好像要说清楚这个题目需要很多字的,我给你说下方法吧.首先把方程里的sinx用cosx代换,然后就把cos3x也用cosx表示(这个太烦了,好像是三次方程,如果不是cos3x而是cosx就好办了),“再
(cosx)^2-sinx+a=01-(sinx)^2-sinx+a=0(sinx+1/2)^2=5/4-asinx=+/-(√(5/4-a))-1/2因为0
∵方程sinx^2-2sinx-a=0有实数解,配方得:(sinx-1)^2=a+1,∵-1≤sinx≤1,∴-2≤sinx-1≤0,∴0≤(sinx-1)^2≤4,即,0≤a+1≤4,∴-1≤a≤3
cos2x-2sinx+2a+1=01-2(sinx)^2-2sinx+2a+1=02(sinx)^2+2sinx=2a+2(sinx)^2+sinx=a+1(sinx)^2+sinx+1/4=a+1
sinx+根号3cosx=2(sinxsinπ/6+cosxcosπ/6)=2cos(x-π/6)所以cos(x-π/6)=-a在(0,π)内有相异两实根这时候x-π/6范围是(-π/6,5π/6)有
∵.a•b=cos2x−sin2x+23sinxcosx=cos2x+3sin2x=2sin(2x+π6)∴sin(2x+π6)=513∵x∈[−π4,π6],∴x∈[−π3,π2]∴cos(2x+π
记t=sinx,则方程化为:2(1-2t^2)+4(a-1)t-4a+1=04t^2-4(a-1)t+4a-3=0因为|sinx|=0得:a>=3+√5,或a
已知x∈[0,2π),分类讨论:1.x∈[0,π),sinx≥0,所以|sinx|=sinx原方程变为cos2x=cos(sinx+sinx)=2cosx*sinx=sin2x即tan2x=1,其中2
1.当a‖b时,-sinx-1.5cosx=0,即tanx=-1.5所以sin2x=2tanx/1+tan^2x=12/13cos2x=1-tan^2x/1+tan^2x=-5/13所以2cos^2x
(1)∵sinx+cosx=a∴a=2sin(x+π4),∴-2≤a≤2(2))∵sinx+cosx=a∴a=2sin(x+π4),设y1=ay2=sin(x+π4),由题意可知y1=ay2=sin(
cos^2x-sin^2x+2sinx+2a-3=01-sin^2x-sin^2x+2sinx+2a-3=01-2sin^2x+2sinx+2a-3=02sin^2x-2sinx-2a+2=0把sin
(1)求曲线y=sinx在点A(π/6,1/2)的切线方程.【解】设f(x)=sinx,则f'(x)=cosx,切线斜率为k=f'(π/6)=(3)/2,切线方程为y=(3/2)[x-(π/6)]+(
化简一下:1-3sin^2x+2sinx+2a+1=03sin^2x-2sinx-2a-2=0sin^2x-2/3*sinx-2/3*(a+1)=0令sinx=T,那么0
sinx=2sinx/2cosx/2=1/3sinx/2cosx/2=1/6(sinx/2+cosx/2)^2=1+2sinx/2cosx/2=4/32π
两边同除以2,得sinx×1/2-√3/2×cosx=(2k-1)/2.因为1/2=sin(π/6),√3/2=cos(π/6),所以原方程化为cos(x+/6)=-(2k-1)/2.x∈[0,π],