已知函数fx=cos²+sinxcosx,求fx单调递增区间

已知函数fx=cos²+sinxcosx,求fx单调递增区间
已知函数fx=cos²+sinxcosx,求fx单调递增区间

已知函数fx=cos²+sinxcosx,求fx单调递增区间
f(x)=cos²x+sinxcosx
=(cos2x+1)/2+1/2sin2x
=(1/2cos2x+1/2sin2x)+1/2
=√2/2*(√2/2cos2x+√2/2sin2x)+1/2
=√2/2*sin(2x+π/4)+1/2
令-π/2+2kπ≤2x+π/4≤π/2+2kπ (k∈Z)
-3π/4+2kπ≤2x≤π/4+2kπ
-3π/8+kπ≤x≤π/8+kπ
所以fx单调递增区间为[-3π/8+kπ,π/8+kπ] (k∈Z)

化简为f(x)=0.5+0.5*[sin(2x+o.25π)]

fx=cos²x+sinxcosx
=1/2(1+cos2x)+1/2sin2x
=1/2sin2x+1/2cos2x+1/2
=√2/2(√2/2sin2x+√2/2cos2x)+1/2
=√2/2sin(2x+π/4)+1/2
由2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z
得kπ-3π/8≤x≤kπ+π/8,k∈Z
∴f(x)单调地鞥区间为[kπ-3π/8,kπ+π/8],k∈Z