函數(shù)
f(x)=(√3)sinxcosx+cos²x+1
=(√3/2)(2sinxcosx)+(1/2)(2cos²x)+1
=(√3/2)sin2x+(1/2)(1+cos2x)+1
=sin(2x)cos(π/6)+cos(2x)sin(π/6)+(3/2)
=sin[2x+(π/6)]+(3/2)
[[[1]]]
易知,
T=π
[[[2]]]
2kπ+(π/2)≤2x+(π/6)≤2kπ+(3π/2)
∴單調(diào)遞減區(qū)間為
kπ+(π/6)≤x≤kπ+(2π/3)
即[kπ+(π/6),kπ+(2π/3)]
[[[3]]]
f(A)=2
sin[2A+30º]+(3/2)=2
sin(2A+30º)=1/2
易知此時(shí)2A+30º=150º
∴A=60º
三角形面積
S=(bc/2)sinA
(c/2)×(√3)/2=√3/2
∴c=2
再由余弦定理可得
a²=b²+c²-2bccosA
=1+4-2=3
∴a=√3