z1=-1+2i
z2=cosa+isina
z1*z2的實(shí)部等于 -cosa-2sina=0
tana=-1/2
tan2a=2tana/[1-tan^2(a)]= - 4/3已知函數(shù)f(x)=2sinxcosx+2cos^2x1）求函數(shù)f(x)的單調(diào)遞增區(qū)間2）將函數(shù)y=f(x)圖像向右平移π/4個(gè)單位后，得到函數(shù)y=g(x)的圖像，求方程g(x)=1的解f(x)=sin2x+(1+cos2x)=sin2x+cos2x+1=√2sin(2x+π/4)+1由-π/2+2kπ≤2x+π/4≤π/2+2kπ得：-3π/8+kπ≤x≤π/8+kπ，所以原函數(shù)的單調(diào)增區(qū)間為：【-3π/8+kπ，π/8+kπ，】2）g(x)=f(-π/4)=√2sin[2(x-π/4)+π/4]+1=√2sin(2x-π/4)+1即g(x)=√2sin(2x-π/4)+1方程g(x)=1==>√2sin(2x-π/4)=02x-π/4=kπx = π/8+kπ/2社關(guān)于x的不等式x(x-a-1)<0(a∈R)的解集為M,不等式（x+1)/(x-3)≤0的解集為N1）當(dāng)a=1時(shí)，求集合M2)若M≦N，求實(shí)數(shù)a的取值范圍1）a=1M=(0,2)N=[-1,3)2) 因?yàn)镸是N的子集，i)當(dāng)M是空集時(shí)，(a+1)^2≤0;==>a=-1ii)當(dāng)M不是空集時(shí)，根據(jù)根的分布M中的兩根在【-1，3）中，<==>{f(-1)>0 {f(3)>0 {Δ>0 {-1<(a+1)/2<3.............................................................................................................................................{a+2>0{6-3a>0{a+1≠0{-3