sin2xcosx

已知sin^2x+sin2xcosx-cos2x=1,0sin^2x这是什么意思

已知x为锐角,且sin^2(2x)+sin2xcosx-cos2x=1求sinx,tanx因为sin^2(2x)+sin2xcosx-cos2x=1cos2x=2cos^2(x)-1所以sin^2(2x)+sin2xcosx-2cos^2(

函数f(x)=(sin2xcosx)/（1-sinx）的值域#答案[-1/2,4)f(x)=sin2xcosx/(1-sinx)=2sinx*(cosx*cosx)/(1-sinx)=2sinx*(1-sinx)*(1+sinx)/(1-s

(sin2x)^2+sin2xcosx-cos2x=1x在0到90度,求sinx,tanx(sin2x)^2+sin2xcosx-cos2x=1[2倍角公式](sin2x)^2+sin2xcosx-[2(cosx)^2-1]=1(sin2x

sin^2(2x)+sin2xcosx-cosx=1x属于(0,π/2)求sinxtanx的值原式化简：（sin2x+cosx+1)(sin2x-1)=0(1)sin2x+cosx+1=02sinxcosx+cosx+1=0因为x属于(0,

1.sin^2(2x)+sin2xcosx-cosx=1x属于(0,π/2)求sinxtanx的值原式化为（sin2x+cosx+1)(sin2x-1)=0(1)sin2x+cosx+1=02sinxcosx+cosx+1=0因为x属于(0

已知sin²2x+sin2xcosx-cos2x=1,x∈(0,2/π）,求sinx,tanx要过程sin2x=2sinxcosx;cos2x+1=2(cosx)^2=>(sin2x)^2+sin2xcosx=cos2x+1=2(

∫x^2/(1-x^2)dx求积分还有个∫1/(sin2xcosx)dx求积分∫x^2/(1-x^2)dx=∫[1/(1-x^2)-1]dx=∫[(1/2)/(1+x)+(1/2)/(1-x)-1]dx=(1/2)ln│(1+x)/(1-x

题目是y=sin平方x乘以cos平方x,进行求导,正确答案是sin2xcosx^2-2xsin平方xsinx^2,我想知道过程,谢谢是高数题目=2sinx(cosx)cosx^2+sinx^22cosx（-sinx）=sin2xcosx^2