10.3 解三角方程式 Solve Trigonometric Equations - 独中课程 | 线上补习

高中一 | 高级数学

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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,

把 $\sqrt{3} cos x + sin x$ 用 $Rcos(x - \alpha)$ 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

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

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

Answer:

(a) 5, -12, 12

(b) -1, 25