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第 18 週 · 三角學(一) · 第 4 節

解三角方程（0°–360°）

解三角方程 = 在 0°–360° 裡找出所有讓等式成立的角。難點是一個方程常有「多個」解 —— 用參考角 + ASTC 把全部象限的解找齊，必要時先換元化成二次。

▶ 第一關： 基本型 sinx=k 兩步解
sin x = 1/2 : line y=1/2 hits the circle at 30 and 150
y = 1/230150
ASTC : sin is + in quadrants I and II → two solutions
ASTCsin +sin +shaded = where sin is positive
build each solution from the reference angle a
x = a180 - a180 + a360 - aa = reference angle (0 to 90)
▶ 第二關： cos、tan 與負值
cos x = 1/2 : cos is + in I and IV → 60 and 300
x = 1/260300
cos x = -1/2 : cos is - in II and III → 120 and 240
x = -1/2120240
tan x = 1 : solutions 45 and 225 (both on the same line)
45225tan repeats every 180
▶ 第三關： 數解的個數（波形）
y = sin x on 0 to 360 ; line y=1/2 crosses twice
y = 1/2301501803602 crossings = 2 solutions
k=1 grazes the peak (1 sol) ; k=2 misses (0 sol)
y = 190 (1 sol)y = 2no crossing -> 0 sol
sin x = 0 on closed 0 to 360 : x = 0, 180, 360 (3 sol)
y = 0 (x-axis)01803603 intersections counted
⚔ BOSS： 換元化二次解方程
substitution t = sin x turns it into a quadratic in t
2 sin²x - sin x - 1 = 0squared=>2t² - t - 1 = 0set t = sin x(2t + 1)(t - 1) = 0 -> t = -1/2 or 1both in [-1, 1] -> keep both
each kept t is solved back to all x in 0 to 360
t = -1/2 or 1sin x = 1sin x = -1/2x = 90x = 210, 330boundary: 1 solIII and IV: 2 solsall solutions: x = 90, 210, 330