Question

已知函数f(x)=2acos^2x+bsinxcosx,f(0)=2,f(π/3)=1/2+（根号3）/2

若f(a)=0,a属于(0,2π),求a的值

Answer 1

f(0)=2,f(π/3)=1/2+（根号3）/2
即：2a=2，
a/2+(√3)b/4=1/2+√3/2
解得：a=1，b=2
所以，f(x)=2cos²x+2sinxcosx
=cos2x+1+sin2x
=√2sin(2x+π/4)+1
f(a)=√2sin(2a+π/4)+1=0
得：sin(2a+π/4)=-√2/2
因为a属于(0,2π)，则可得：a1=3π/4，a2=7π/4，a3=π/2，a4=3π/2

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Answer 2

f(x)=2acos^2x+bsinxcosx
=a(1+cos2x)+1/2bsin2x
=1/2bsin2x+acos2x+a
f(0)=1/2bsin0+acos0+a=2, 得 a=1,
f(π/3)=1/2bsin2π/3+acos2π/3+a=1/2+√3/2
√3/4b-1/2*a+a=1/2+√3/2,得 b=2
f(x)=sin2x+os2x+1=√2sin(2x+π/4)+1
f(a)=√2sin(2a+π/4)+1=0
sin(2a+π/4)= -√2/2
a属于(0,2π),
2a+π/4=5π/4, 或 2a+π/4=7π/4
a=π/2 或 a=3π/4