cos3xcosx=-sinx(sin3x+sinx) 0≤x≤pai 求x

2025-01-21 02:58:12
推荐回答(2个)
回答1:

cos3xcosx=-sinx(sin3x+sinx)
cos3xcosx+sin3xsinx+sin^2 x=0
cos(3x-x)+sin^2 x=0
cos2x+sin^2 x=0
1-2sin^2 x+sin^2 x=0
sin^2 x=1
0≤x≤pai
则sinx>0
所以
sinx=1
x=pai/2

回答2:

cos3xcosx+sinxsin3x=-(sinx)^2,则cos2x+(1-cos2x)/2=0,co2x=-1,2x=2Kπ+π
所以x=Kπ+π/2,K∈Z.