第四章:部分分式 Partial Fraction

第六章:角的形成及单位 Angles and Measurements

10.3 解三角方程式 Solve Trigonometric Equations
Solve the following trigonometric equations with given range of angle (1 ~ 11)
在所给的角度范围解下列各三角方程式(1 ~ 11)

(1) 3 - 3 cos x = 2 sin^2x       0^{\circ} \leq x \leq 360^{\circ}

Answer:
0^{\circ}, 60^{\circ}, 300^{\circ}, 360^{\circ}

(2) 3 tan^2x + 5 = 7 sec x        -180^{\circ} \leq x \leq 180^{\circ}

Answer:

-60^{\circ}, 60^{\circ}

(3) sin x + 2 cos x = 0        -180^{\circ} \leq x \leq 180^{\circ}

Answer:

-63.43^{\circ}, 116.57^{\circ}

(4) 6 sin^2x = 8 sin x cos x- 1    0^{\circ} \leq x \leq 180^{\circ}

Answer:
8.13^{\circ}, 45^{\circ}
(5) sin x cos x = \frac{\sqrt2}{4}    -180^{\circ} \leq x \leq 180^{\circ}
Answer:
22.5^{\circ}, 67.5^{\circ}, -112.5^{\circ}, -157.5^{\circ}
(6) 112 sin x + cos x = cos3x    0 \leq x \leq 2\pi
Answer:
0, \pi, 2\pi, \frac{3}{4}\pi, 1\frac{3}{4}\pi
(7) 3 sin x - 4 cos x = 2 - 3 sin2x    0^{\circ} \leq x \leq 360^{\circ}
Answer:
41.81^{\circ}, 120^{\circ}, 138.19^{\circ} , 240^{\circ}
(8) sin2 \frac{1}{2}x = sin^2x    -180^{\circ} \leq x \leq 180^{\circ}
Answer:
0^{\circ}, 120^{\circ}, -120^{\circ}
(9) 2 cos2x = cos x- sin x    0^{\circ} \leq x \leq 360^{\circ}
Answer:
45^{\circ}, 114.3^{\circ}, 225^{\circ}, 335.7^{\circ}
(10) sin3x = cos x    0 \leq x \leq 2\pi
Answer:
\frac{1}{4}\pi , 1\frac{1}{4}\pi , \frac{1}{8}\pi , \frac{5}{8}\pi , 1\frac{1}{8}\pi , 1\frac{5}{8}\pi
(11) sin2x + sin x + 2cos x + 1 = 0    0 \leq x \leq 2\pi
Answer:
\frac{2}{3}\pi , \frac{4}{3}\pi , \frac{3}{2}\pi
Give the general solution of the following trigonometric equations. (12 ~ 17)
求下列各三角方程式的一般解(12 ~ 17)
(12) sin x + sin2x + sin3x = 0   
Answer:
\frac{1}{2}n\pi , 2n\pi\pm\frac{2}{3}\pi
(13) (cosec2x - tan x )^2 = 3   
Answer:
\frac{n\pi}{2}\pm\frac{\pi}{12}
(14) cos^22x = cos^2x   
Answer:
\frac{1}{3}n
(15) \frac{1-sin{x}}{{cos}^2{x}}=2-cos{x}-sin{x}   
Answer:
n\pi + \frac{\pi}{4}, 2n\pi
(16) tan3x = tan x   
Answer:
\frac{1}{2}n\pi
(17) sin3x = sin2x   
Answer:
2n\pi , \frac{2n+1}{5}\pi
Solve the following a sin x + b cos x = c trigonometric equations with given range of angle (18 ~ 21)
在所给的角度范围解下列a sin x + b cos x = c 的三角方程式(18 ~ 21)

(18) 3 cos x + 5 sin x = 2     0^{\circ} \leq x \leq 360^{\circ}

Answer:
128.98^{\circ}, 349.10^{\circ}

(19) \sqrt3cos x- sin x = 1     0 \leq x \leq 2\pi

Answer:
\frac{\pi}{6},\ \frac{3\pi}{2}

(20) sin2x + 2cos2x = 1     0^{\circ} \leq x \leq 360^{\circ}

Answer:

45^{\circ}, 161.57^{\circ}, 225^{\circ}, 341.57^{\circ}

(21) sin2x - cos2x = \sqrt2
(一般解、general solution)
Answer:
n\pi + \frac{3}{8}\pi
(22) Express \sqrt{3} cos x + sin x in term of Rcos(x - \alpha) where R > 0 and \alpha is acute angle. Find the values of R and \alpha. Hence,

\sqrt{3} cos x + sin xRcos(x - \alpha) 表示,其中 R > 0 且 \alpha 为一锐角. 求 R 和 \alpha 的值。据此

(a) write down the maximum value of \sqrt{3}cos x + sin x

写下\sqrt{3}cos x + sin x的最大值

(b) solve the equation \sqrt{3}cos x + sin x = 1, where 0 \leq x \leq 2\pi

解方程式 \sqrt{3}cos x + sin x = 1, 其中 0 \leq x \leq 2\pi

Answer:
(a) 2

(b) \frac{1}{2}\pi , \frac{11\pi}{6}

(23) If f(x) = 17cos^2x - 24sin x cos x + 7sin^2 x = a cos2x + b sin2x + c, find

f(x) = 17cos^2x - 24sin x cos x + 7sin^2 x = a cos2x + b sin2x + c, 求

(a) the values of a, b and c,

a, b 和 c的值,

(b) the maximum and minimum values of f (x),

f (x) 的最大和最小值

(c) the general solution of the equation 17cos^2x - 24sin x cos x + 7sin^2x = 15.

17cos^2x - 24sin x cos x + 7sin^2x = 15的一般解

Answer:
(a) 5, -12, 12

(b) -1, 25

(c) n·180^{\circ} + 4.64^{\circ}, n·180^{\circ} – 72.02^{\circ}