若方程cosx-sin2x-a=0有解
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2sin2x+cos2x=1,所以4sinx*cosx=2(sinx)^2得出x=0或则tanx=2,当x=0时,[2(cosx)^2+sin2x]/(1+tanx)=2当tanx=2时,[2(cos
将原方程转化为:a=-sin2x-cosx=cos2x-cosx-1,设cosx=t,t∈[-1,1],则a=t2-t-1=(t-12)2-54∈[-54,1]故答案为:[-54,1];
原方程可化为cos2x-2cosx-a-1=0,令t=cosx,得t2-2t-a-1=0,原问题转化为方程t2-2t-a-1=0在〔-1,1〕上至少有一个实根.令f(t)=t2-2t-a-1,对称轴t
f(x)=1+sin2x+sin²x-cos²x=sin2x-cos2x+1=(根号2)sin(2x-π/4)+1f(θ)=(根号2)sin(2θ-π/4)+1=8/5所以sin(
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sin2x/sinx=cos2x/cosx两边除以:cos2x/cosxtan2x/tanx=1tan2x=tanx2tanx/(1+(tanx)^2)=tanx(tanx)^2=1tanx=1或ta
sin2x+2sinx^2+sinx+cosx=02sinxcosx+2sinxsinx+sinx+cosx=02sinx(sinx+cosx)+sinx+cosx=0(2sinx+1)(sinx+c
f(x)=2cos^2x+√3sin2x=1+cos2x+√3sin2x=2sin(2x+π/6)+1(1)若f(x)=0,sin(2x+π/6)=-1/2x∈(-π/2,0),x=-π/62x=-π
首先对y=sin2x-2(sinx+cosx)+a²进行化简采用参数变换,令t=sinx+cosx=√2sin(x+pi/4),易知,t的范围为[-√2,√2]t^2=1+2sinx*cos
f(x)=向量a.向量b.=(1+sin2x)*1+(sinx-cosx)*(sinx+cosx).=1+sin2x-(cos^2x-sin^2x).=1+sin2x-cos2x.=1+√2sin(2
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原式=f[cos(π/2-x)]=2-sin[2(π/2-x)]=2-sin(π-2x)=2-sin2x选C
令a=sinx,b=cosx,有a^2+b^2=1方程化为:1+a+b+2ab+b^2-a^2=0上两式相加得:2b^2+a+b+2ab=0即2b(b+a)+(a+b)=0(a+b)(2b+1)=0得
sinx×cos2x-sin2x×cosx=sin(x-2x)=-sinx
程4cosx+sin2x+m-4=0可化为m=4-4cosx-sin2x=cos2x-4cosx+3=(cosx-2)2-1∵cosx∈[-1,1],则=(cosx-2)2-1∈[0,8]则若关于x的
sin2x-12(sinx-cosx)+12=0-(sinx-cosx)^2+1-12(sinx-cosx)+12=o(sinx-cosx)^2+12(sinx-cosx)+13=0设sinx-cos
第一问很好解(a+b)=(cos2x+cosx,sin2x+sinx)a-b=(cos2x-cosx,sin2x-sinx)所以相乘得cos2x^2-cosx^2+sin2x^2-sinx^2=1-1
向量AC=(cosx-3,sinx)向量BC=(cosx,sinx-3)向量AC.向量BC=cosx(cosx-3)+sinx(sinx-3)=-1cos²x+sin²x-3(si
由原方程得:3(sinx-cosx-1)=sin2x-1+1右边=2sinx·cosx-(sinx^2+cosx^2)+1=1-(sinx-cosx)^2令sinx-cosx=y则有3(y-1)=1-