Course: 14.11 使用四方技术解决三角等同

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使用二次曲线技术解决三角等量
As Agent Trigonometry, you are given this clue: $\begin{aligned} 2 \cos^{2} x - 3 \cos x + 1 = 0 \end{aligned}$ . If $\begin{aligned} x \end{aligned}$ falls in the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ , what is/are its possible value(s)?
::作为代理 Trigonology , 您获得了这个线索 : 2cos2 x-3cosx+1=0。如果 x 坠落于 0. x < 2 的间距, 那么它可能值是多少 ?

Solving Trigonometric Equations
::解决三角等量等量

Another way to solve a trig equation is to use factoring or the quadratic formula. Let’s look at a few problems .
::另一种解决三重方程的方法是使用保理或二次公式。让我们来看看几个问题。

Let's solve the following trigonometric equations.
::让我们解决以下三角方程。

Solve $\begin{aligned} \sin^{2} x - 3 \sin x + 2 = 0 \end{aligned}$ .
::解决sin2x-3sinx+2=0。

This sine equation looks a lot like the quadratic $\begin{aligned} x^{2} - 3 x + 2 = 0 \end{aligned}$ which factors to be $\begin{aligned} (x - 2) (x - 1) = 0 \end{aligned}$ and the solutions are $\begin{aligned} x = 2 \end{aligned}$ and 1. We can factor the trig equation in the exact same manner. Instead of just $\begin{aligned} x \end{aligned}$ , we will have $\begin{aligned} \sin x \end{aligned}$ in the factors.
::这个正弦方程看起来与四方形 x2 - 3x+2=0( x-2 (x-1)=0) 的因数非常相似, 答案是 x=2 和 1 。我们可以以同样的方式乘以三方方方程。与其只乘以 x, 我们的因数中将会有 sinx 。

$\begin{aligned} \sin^{2} x - 3 \sin x + 2 & = 0 \\ (\sin x - 2) (\sin x - 1) & = 0 \\ \sin x = 2 a n d \sin x & = 1 \end{aligned}$

:sinx-2)(sinx-1)=0sinx=2和sinx=1)

There is no solution for $\begin{aligned} \sin x = 2 \end{aligned}$ and $\begin{aligned} \sin x = 1 \end{aligned}$ when $\begin{aligned} x = \frac{π}{2} \pm 2 π n \end{aligned}$ .
::当 x222n 时, sin2=2 和 sinx=1 没有解决之道。

Solve $\begin{aligned} 1 - \sin x = \sqrt{3} \cos x \end{aligned}$ in the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::在 0 x < 2 间距内, 溶解 1 - sinx= 3cosx 。

To solve this equation, use the Pythagorean Identity $\begin{aligned} \sin^{2} x + \cos^{2} x = 1 \end{aligned}$ . Solve for either cosine and substitute into the equation. $\begin{aligned} \cos x = \sqrt{1 - \sin^{2} x} \end{aligned}$
::要解析此方程式, 请使用 Pythagorea 身份身份 sin2\\\ x+cos2\\\\\ x=1. 解决两个余弦并替换为公式。 cos=x=1- sin2\\\ x

$\begin{aligned} 1 - \sin x & = \sqrt{3} \cdot \sqrt{1 - \sin^{2} x} \\ (1 - \sin x)^{2} & = {\sqrt{3 - 3 \sin^{2} x}}^{2} \\ 1 - 2 \sin x + \sin^{2} x & = 3 - 3 \sin^{2} x \\ 4 \sin^{2} x - 2 \sin x - 2 & = 0 \\ 2 \sin^{2} x - \sin x - 1 & = 0 \\ (2 \sin x + 1) (\sin x - 1) & = 0 \end{aligned}$

::1- 辛x=3- 3- 辛x2x( 1- 辛x)2=3- 3sin2x21-2sinx+sin2x=3- 3sin2x4sin2x-2x-2sinx-2=02sin2x- 辛x-1=0( 2sinx+1)(sinx-1)=0

Solving each factor for $\begin{aligned} x \end{aligned}$ , we get $\begin{aligned} \sin x = - \frac{1}{2} \to x = \frac{7 π}{6} \end{aligned}$ and $\begin{aligned} \frac{11 π}{6} \end{aligned}$ and $\begin{aligned} \sin x = 1 \to x = \frac{π}{2} \end{aligned}$ .
::解决 x 的每个因数时, 我们得到sinx= 12x= 76 和 116 和 sinx= 1x 。

Solve $\begin{aligned} \tan^{2} x - 5 \tan x - 9 = 0 \end{aligned}$ in the interval $\begin{aligned} 0 \leq x < π \end{aligned}$ .
::在间距 0 x 中解决 tan2 x- 5tanx- 9= 0 。

This equation is not factorable so you have to use the Quadratic Formula.
::此方程式不可计算, 所以您必须使用二次曲线公式。

$\begin{aligned} \tan x & = \frac{5 \pm \sqrt{{(- 5)}^{2} - 4 (1) (- 9)}}{2} \\ = \frac{5 \pm \sqrt{61}}{2} \\ \approx 6.41 a n d - 1.41 \end{aligned}$

::x=5(-5)2-4(1)(-(9)2=56126.41和-1.41

$\begin{aligned} x \approx \tan^{- 1} 6.41 \approx 1.416 rad \end{aligned}$ and $\begin{aligned} x \approx \tan^{- 1} - 1.41 \approx - 0.954 rad \end{aligned}$
::Xátan-116.411.416 rad和 xtan-1-1-1.410.954 rad

The first answer is within the range, but the second is not. To adjust -0.954 to be within the range, we need to find the answer in the second quadrant, $\begin{aligned} π - 0.954 = 2.186 rad \end{aligned}$ .
::第一个答案在范围以内,第二个答案不是。要调整 -0.954 以在范围以内,我们需要在第二个象限中找到答案,= 0.954= 2.186 rad。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the possible values for $\begin{aligned} 2 \cos^{2} x - 3 \cos x + 1 = 0 \end{aligned}$ .
::早些时候, 您被要求找到 2cos2 的值, x- 3cosx+1=0 。

We can solve this problem by factoring.
::我们可以通过保理来解决这个问题。

$\begin{aligned} 2 \cos^{2} x - 3 \cos x + 1 = 0 \\ (2 \cos x - 1) (\cos x - 1) = 0 \\ \cos x = \frac{1}{2} O R x = 1 \end{aligned}$

::2cos2_x- 3cos_ x+1=0( 2cos_ x-1)( os_ x-1) =0cos_ x=12ORx=1

Over the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ , $\begin{aligned} \cos x = \frac{1}{2} \end{aligned}$ when $\begin{aligned} x = \frac{π}{3} \end{aligned}$ and $\begin{aligned} \cos x = 1 \end{aligned}$ when $\begin{aligned} x = 0 \end{aligned}$ .
::时间间隔为 0x < 2, 当 x3 和 osx=1 时, COsx=12 x=0 时。

Solve the following trig equations using any method in the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::使用间隔 0x<2中的任何方法解决以下三重方程。

Example 2
::例2

$\begin{aligned} \sin^{2} x \cos x = \cos x \end{aligned}$
:xcosx=cosxx)

Put everything onto one side of the equation and factor out a cosine.
::把所有东西放在方程的一边分出一个连弦

$\begin{aligned} \sin^{2} x \cos x - \cos x & = 0 \\ \cos x (\sin^{2} x - 1) & = 0 \\ \cos x (\sin x - 1) (\sin x + 1) & = 0 \end{aligned}$

::sin2_x_xx_1(sin_x+1)=0x(sin2_x_1)=0x(sin_x+1)=0x(sin_x+1)=0x(sin_x+1)

$\begin{aligned} \cos x & = 0 \sin x = 1 \sin x = - 1 \\ x & = \frac{π}{2} a n d \frac{3 π}{2} x = \frac{π}{2} x = \frac{3 π}{2} \end{aligned}$

::cosx=0 sinx=1 sinx*1x2 和 32 x2x=32

Example 3
::例3

$\begin{aligned} \sin^{2} x = 2 \sin (- x) + 1 \end{aligned}$
::sin2\\ xx=2sin}(- x)+1

Recall that $\begin{aligned} \sin (- x) = - \sin x \end{aligned}$ from the Negative Angle Identities.
::回顾从负角辨别物中得出的sin(-x)sinx(sinx) 。

$\begin{aligned} \sin^{2} x & = 2 \sin (- x) + 1 \\ \sin^{2} x & = - 2 \sin x + 1 \\ \sin^{2} x + 2 \sin x + 1 & = 0 \\ (\sin x + 1)^{2} & = 0 \\ \sin x & = - 1 \\ x & = \frac{3 π}{2} \end{aligned}$

::sin2\\ xx=2sin}(- x)+1sin2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Example 4
::例4

$\begin{aligned} 4 \cos^{2} x - 2 \cos x - 1 = 0 \end{aligned}$
::4cos2x-2cosx-1=0

This quadratic is not factorable, so use the quadratic formula.
::此二次方程式不可乘以, 所以使用二次方程式。

$\begin{aligned} \cos x & = \frac{2 \pm \sqrt{2^{2} - 4 (4) (- 1)}}{2 (4)} \\ = \frac{2 \pm \sqrt{20}}{8} \\ = \frac{1 \pm 2 \sqrt{5}}{4} \end{aligned}$

::COsx=222-4(4)(-1)2(4)=2208=1254

$\begin{aligned} x & \approx \cos^{- 1} (\frac{1 + \sqrt{5}}{4}) & x \approx \cos^{- 1} (\frac{1 - \sqrt{5}}{4}) \\ \approx \cos^{- 1} 0.8090 a n d & \approx \cos^{- 1} - 0.3090 \\ \approx 0.6283 & \approx 1.8850 (reference angle is π - 1.8850 \approx 1.2570) \end{aligned}$

::Xcos-11(1+54xcos-11(1-54)cos-10.8090和cos-1-0.3090+0.6283)1.8850(参考角为1.8850_1.2570)

The other solutions in the range are $\begin{aligned} x \approx 2 π - 0.6283 \approx 5.6549 \end{aligned}$ and $\begin{aligned} x \approx π + 1.2570 \approx 4.3982 \end{aligned}$ .
::其他的解决方案是 x20.62835.6549 和 x1.25704.3982。

Review
::回顾

Solve the following trig equations using any method. Find all solutions in the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ . Round any decimal answers to 4 decimal places.
::使用任何方法解决以下三重方程。在 0. x < 2Q 间距中查找所有解决方案。将小数点后的任何答案四舍五入到小数点后四位。

$\begin{aligned} 2 \cos^{2} x - \sin x - 1 = 0 \end{aligned}$
::2cos2x-sinx-1=0

$\begin{aligned} 4 \sin^{2} x + 5 \sin x + 1 = 0 \end{aligned}$
::4sin2x+5sinx+1=0

$\begin{aligned} 3 \tan^{2} x - \tan x = 0 \end{aligned}$
::3tan2x-tanx=0

$\begin{aligned} 2 \cos^{2} x + \cos (- x) - 1 = 0 \end{aligned}$
::2cos2x+cos(- x)- 1=0

$\begin{aligned} 1 - \sin x = \sqrt{2} \cos x \end{aligned}$
::1 - 辛x=2cosx

$\begin{aligned} \sqrt{\sin x} = 2 \sin x - 1 \end{aligned}$
::exix=2sinx-1

$\begin{aligned} \sin^{3} x - \sin x = 0 \end{aligned}$
::sin3 -x -sin*x=0

$\begin{aligned} \tan^{2} x - 8 \tan x + 7 = 0 \end{aligned}$
::tan2x-8tanx+7=0

$\begin{aligned} 5 \cos^{2} x + 3 \cos x - 2 = 0 \end{aligned}$
::5cos2x+3cosx-2=0

$\begin{aligned} \sin x - \sin x \cos^{2} x = 1 \end{aligned}$
::excos2\\ x=1 内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内,内

$\begin{aligned} \cos^{2} x - 3 \cos x + 2 = 0 \end{aligned}$
::COs2%%x- 3cos%x+2=0

$\begin{aligned} \sin^{2} x \cos x = 4 \cos x \end{aligned}$
:xcosx=4cosx)

$\begin{aligned} \cos x \csc^{2} x + 2 \cos x = 6 \cos x \end{aligned}$
::COSxcsc2x+2cosx=6cosx

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval $\begin{aligned} 0 \leq x < 2 π \end{aligned}$ .
::使用您的图形计算计算器,绘制以下方程式,并确定间距 0x < 2 的交叉点。

$\begin{aligned} y & = \sin^{2} x \\ y & = 3 \sin x - 1 \end{aligned}$

::y=sin2xy=3sin*x-1

$\begin{aligned} y & = 4 \cos x - 3 \\ y & = - 2 \tan x \end{aligned}$

::y=4cos_x_3y_2tan_x

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

使用二次曲线技术解决三角等量

Solving Trigonometric Equations ::解决三角等量等量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Solving Trigonometric Equations
::解决三角等量等量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)