已知方程 asinx+bcosx=c。①其中a、b、c都是给定的实数,且a、b不同时为零,x∈[x_0,x_0+2π),x_0是任一固定常数。设△=a~2+b~2-c~2,则当△>0时,方程①有两个不相等的实数根; 当△=0时,方程①有两个相等的实数根; 当△<0时,方程①没有实数根; 证明∵a、b不同时为零, ∴(a~2+b~2)~(1/2)≠0。∴sin(x+φ)=C/((a~2+b~2)~(1/2))。②(其中φ是辅助角,a≠0时,tgφ=b/a;b≠0 The known equation asinx+bcosx=c. 1 where a, b, and c are all given real numbers, and a and b are not zero at the same time, x∈[x_0, x_0+2π), and x_0 is any fixed constant. Let △ = a ~ 2+ b ~ 2-c ~ 2, then when △> 0, Equation 1 has two unequal real roots; when △ = 0, Equation 1 has two equal real roots; When △<0, equation 1 has no real roots; prove that ∵a, b are not zero at the same time, and ∴(a~2+b~2)~(1/2)≠0. ∴sin(x+φ)=C/((a~2+b~2)~(1/2)). 2 (where φ is the auxiliary angle, a≠0, tgφ=b/a; b≠0