f(θ)=(cosθ)^2+[cos(θ+a)]^2+[cos(θ+b)]^2
=(1+cos2θ)/2+[1+cos(2θ+2a)]/2+[1+cos(2θ+2b)]/2
=3/2+[cos2θ+cos(2θ+2a)+cos(2θ+2b)]/2
=3/2+[cos2θ+cos2θcos2a-sin2θsin2a+cos2θcos2b-sin2θsin2b]/2
=3/2+[cos2θ(1+cos2a+cos2b)-(sin2a+sin2b)]/2
如果f(θ)與θ無關(guān),
1+cos2a+cos2b=0.1
sin2a+sin2b=0.2
由1式得
2cos(a+b)cos(a-b)=-1.3
由1式得
2sin(a+b)cos(a-b)=0.4
顯然sin(a+b)=0,cos(a-b)=±1/2.
a+b=nπ,a-b=kπ/2±π/6.
因此a、b存在,如：a=2π/3,b=π/3.