各位會不會覺得題目怪怪的?
這題x有無限多解
a,b是無限多解中之相異兩個??
應該是沒設 x 的範圍吧! 如果有設 0 ≦ x < 2π, 就是兩個解了! 如圖
3cos(x) + 4sin(x) = 2
設 s = sin(x)
3√(1-s2) + 4s = 2
4s - 2 = - 3√(1-s2)
(4s - 2)2 = (- 3√(1-s2))2
25s2 - 16s - 5 = 0
s = (8-3√21)/25, (8+3√21)/25 = sin(a), sin(b) ..
cos(a) = √(1-sin2(a))
= √(1-((8-3√21)/25)^2)
= (4√21 +6)/25
cos(b) = √(1-sin2(b))
= √(1-((8+3√21)/25)^2)
= (4√21 -6)/25
sin(a+b) = sin(a)cos(b) + sin(b)cos(a)
= ((8-3√21)/25)((4√21 -6)/25) + ((8+3√21)/25)((4√21 +6)/25)
= 4√21/25
cos(a+b) = cos(a)cos(b) - sin(a)sin(b)
= ((4√21 +6)/25)((4√21 -6)/25) + ((8-3√21)/25)((8+3√21)/25)
= 7/25
tan(t/2) = (1 - cos(t)) /sin(t)
tan((a+b)/2) = (1 - cos(a+b)) / sin(a+b)
= (1 - (7/25)) / (4√21/25)
= 3·√21 / 14