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新编高中《代数》上册第297页给出了三角方程asinx+bcosx=c有解的充要条件|c/(a~2+b~2)~(1/2)|≤1,进一步探讨可知:对于方程asinx+bcosx=c(0≤x<2π),记△=a~2+b~2-c~2,则(1)方程有两个不同的解△>0;(2)方程有唯一解△=0;(3)方程无解△<0.我们引出符号△=a~2+b~2-c~2,称为三角方程asinx+bcosx=c的判别式.容易证明:方程asinx+bcosx=c.①其中a、b、C都是给定的实
New High School “Algebra” on the first page on page 297 given the trigonometric equation asinx + bcosx = c have the necessary and sufficient conditions | c / (a ~ 2 + b ~ 2) ~ (1/2) | ≤ 1, to further explore It can be seen that (1) the equation has two different solutions △> 0 for the equation asinx + bcosx = c (0≤x <2π) and Δ = a ~ 2 + b ~ 2-c ~ ) Equation has a unique solution △ = 0; (3) the equation has no solution △ <0. We draw the symbol △ = a ~ 2 + b ~ 2-c ~ 2, called the trigonometric equation asinx + bcosx = c discriminant It is easy to prove that the equation asinx + bcosx = c.① where a, b, C are given real