Course: 14.10 用代数解决三角等同

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使用代数解决三角等量
G iven: $\begin{align*}2 \sin x-\sqrt{2}=0\end{align*}$ . If $\begin{align*}0 \le x < 2 \pi\end{align*}$ , what is/are the value(s) of $\begin{align*}x\end{align*}$ ?
::给定值: 2sinx-2=0。如果 0x < 2, x 的值是多少?

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

We have already verified trigonometric identities, which are true for every real value of $\begin{align*}x\end{align*}$ . In this concept, we will solve trigonometric equations. An equation is only true for some values of $\begin{align*}x\end{align*}$ .
::我们已经核实了三角特征, 这对x的每一个真实值都是真实的。在这个概念中, 我们将解决三角方程。公式只对x的某些值是真实的。

Let's verify that $\begin{align*}\csc x-2=0\end{align*}$ when $\begin{align*}x=\frac{5 \pi}{6}\end{align*}$ .
::让我们验证 x=5=6 时 cscx- 2=0 。

Substitute in $\begin{align*}x=\frac{5 \pi}{6}\end{align*}$ to see if the equations holds true.
::以 x= 56 代替来查看方程式是否正确。

$\begin{aligned} \csc (\frac{5 π}{6}) - 2 & = 0 \\ \frac{1}{\sin (\frac{5 π}{6})} - 2 & = 0 \\ \frac{1}{\frac{1}{2}} - 2 & = 0 \\ 2 - 2 & = 0 \end{aligned}$
$\begin{align*}\csc \left(\frac{5 \pi}{6}\right)-2&=0 \\ \frac{1}{\sin \left(\frac{5 \pi}{6}\right)}-2&=0 \\ \frac{1}{\frac{1}{2}}-2&=0 \\ 2-2&=0\end{align*}$
::csc(56)-2=01sin(56)-2=0112-2=02-2=0

This is a true statement, so $\begin{align*}x=\frac{5 \pi}{6}\end{align*}$ is a solution to the equation.
::这是一个真实的语句, 所以 x=56 是方程式的解决方案。

Now, let's solve $\begin{align*}2 \cos x+1=0\end{align*}$ .
::现在,让我们解决2cosx+1=0。

To solve this equation, we need to isolate $\begin{align*}\cos x\end{align*}$ and then use inverse to find the values of $\begin{align*}x\end{align*}$ when the equation is valid.
::要解析此方程式, 我们需要分离 cosx, 然后在公式有效时使用反向来找到 x 的值。

$\begin{aligned} 2 \cos x + 1 & = 0 \\ 2 \cos x & = - 1 \\ \cos x & = - \frac{1}{2} \end{aligned}$
$\begin{align*}2 \cos x+1&=0 \\ 2 \cos x&=-1 \\ \cos x&=- \frac{1}{2}\end{align*}$
::2cos*x+1=02cos*x*1cos*x*12

So, when is the $\begin{align*}\cos x=- \frac{1}{2}\end{align*}$ ? Between $\begin{align*}0 \le x< 2 \pi, x=\frac{2 \pi}{3}\end{align*}$ and $\begin{align*}\frac{4 \pi}{3}\end{align*}$ . But, the trig functions are periodic, so there are more solutions than just these two. You can write the general solutions as $\begin{align*}x=\frac{2 \pi}{3} \pm 2 \pi n\end{align*}$ and $\begin{align*}x=\frac{4 \pi}{3} \pm 2 \pi n\end{align*}$ , where $\begin{align*}n\end{align*}$ is any integer. You can check your answer graphically by graphing $\begin{align*}y=\cos x\end{align*}$ and $\begin{align*}y=- \frac{1}{2}\end{align*}$ on the same set of axes. Where the two lines intersect are the solutions.
::那么, COsx* 12 是什么时候? 在 0x < 2, x=23 和 43 之间 ? 但是, 三角函数是周期性的, 所以有比这两个函数更多的解决方案。您可以以 x=232n 和 x=432n 写入一般解决方案, n 是任意的整数。您可以用图形在相同的轴上绘制 y=cosx 和 y12 来查看您的答案。两条线是相交的解决方案。

Finally, let's solve $\begin{align*}5 \tan(x+2)-1=0\end{align*}$ , where $\begin{align*}0 \le x < 2 \pi\end{align*}$ .
::最后,让我们解决5tan(x+2)-1=0,0x<2。

In this problem, we have an interval where we want to find $\begin{align*}x\end{align*}$ . Therefore, at the end of the problem, we will need to add or subtract $\begin{align*}\pi\end{align*}$ , the period of tangent, to find the correct solutions within our interval.
::在此问题上,我们有一个间隔,我们想找到x。因此,在问题结束时,我们需要增加或减去相左时期,以便在我们的间隔内找到正确的解决办法。

$\begin{aligned} 5 \tan (x + 2) - 1 & = 0 \\ 5 \tan (x + 2) & = 1 \\ \tan (x + 2) & = \frac{1}{5} \end{aligned}$
$\begin{align*}5 \tan(x+2)-1&=0 \\ 5 \tan(x+2)&=1 \\ \tan(x+2)&=\frac{1}{5}\end{align*}$
::5tan(x+2)-1=05tan(x+2)=1tan(x+2)=15

Using the $\begin{align*}\tan^{-1}\end{align*}$ button on your calculator, we get that $\begin{align*}\tan^{-1} \left(\frac{1}{5}\right)=0.1974\end{align*}$ . Therefore, we have:
::使用您的计算器上的 tan-1 按钮, 我们得到的 tan-1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ -\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\"\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

$\begin{aligned} x + 2 & = 0.1974 \\ x & = - 1.8026 \end{aligned}$
$\begin{align*}x+2&=0.1974 \\ x&=-1.8026\end{align*}$
::x+2 =0. 1974x @ @ @% 1. 8026

This answer is not within our interval. To find the solutions in the interval, add $\begin{align*}\pi\end{align*}$ a couple of times until we have found all of the solutions in $\begin{align*}[0, 2 \pi]\end{align*}$ .
::此答案不在我们的间隔内。要找到间隔内的解决方案, 请在找到 [ 02] 中的所有解决方案之前, 添加 + 几次。

$\begin{aligned} x & = - 1.8026 + π = 1.3390 \\ = 1.3390 + π = 4.4806 \end{aligned}$
$\begin{align*}x&=-1.8026+ \pi=1.3390 \\ &=1.3390+ \pi=4.4806\end{align*}$
::1.80261.3390=1.33904.4806

The two solutions are $\begin{align*}x = 1.3390\end{align*}$ and 4.4806.
::两种解决办法是x=1.3390和4.4806。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the value of x from the equation $\begin{align*}2 \sin x-\sqrt{2}=0\end{align*}$ .
::早些时候, 您被要求从 2sinx-2=0 方程式中找到 x 的值。

To solve this equation, we need to isolate $\begin{align*}\sin x\end{align*}$ and then use inverse to find the values of $\begin{align*}x\end{align*}$ when the equation is valid.
::要解决这个方程式, 我们需要分离 sinx , 然后在公式有效时使用反向来找到 x 的值。

$\begin{aligned} 2 \sin x - \sqrt{2} = 0 \\ 2 \sin x & = \sqrt{2} \\ \sin x & = \frac{\sqrt{2}}{2} \end{aligned}$
$\begin{align*}2 \sin x -\sqrt{2}=0 \\ 2 \sin x&=\sqrt{2} \\ \sin x&=\frac{\sqrt{2}}{2}\end{align*}$
::2sinx-2=02sinx=2sinx=22

So now we need to find the values of $\begin{align*}x\end{align*}$ for which $\begin{align*}\sin x =\frac{\sqrt{2}}{2}\end{align*}$ . We know from the special triangles that this value of sine holds true for a $\begin{align*}45^\circ\end{align*}$ angle, but is that the only value of $\begin{align*}x\end{align*}$ for which it is true?
::所以我们现在需要找到 x 的值, 因为 sinx=22 。我们从特殊的三角中知道, 这个正弦值对于 45 角度来说是真实的, 但是它是 x 唯一真实的值吗 ?

We are told that $\begin{align*}0 \le x < 2 \pi\end{align*}$ . Recall that the sine is positive in both the first and second quadrants, so $\begin{align*}\sin x =\frac{\sqrt{2}}{2}\end{align*}$ when $\begin{align*}x\end{align*}$ also is $\begin{align*}135^\circ\end{align*}$ .
::我们被告知 0x<2Q。回顾正弦在第一和第二四点都是正的, 所以sinx=22 当 x 也为 135 时。

Example 2
::例2

Determine if $\begin{align*}x=\frac{\pi}{3}\end{align*}$ is a solution for $\begin{align*}2 \sin x=\sqrt{3}\end{align*}$ .
::确定 x3 是否为 2sinx=3 的解决方案。

$\begin{align*}2 \sin \frac{\pi}{3}= \sqrt{3} \rightarrow 2 \cdot \frac{\sqrt{3}}{2}=\sqrt{3}\end{align*}$ Yes, $\begin{align*}x=\frac{\pi}{3}\end{align*}$ is a solution.
::2sin3=3232=3 是, x3是一个解决方案。

Example 3
::例3

Solve the following trig equation in the interval $\begin{align*}0 \le x< 2 \pi\end{align*}$ .
::在 0. x < 2 间距内解决以下三重方程。

$\begin{align*}9 \cos^2x-5=0\end{align*}$
::9cos2x-5=0

$\begin{aligned} 9 \cos^{2} x - 5 & = 0 \\ 9 \cos^{2} x & = 5 \\ \cos^{2} x & = \frac{5}{9} \\ \cos x & = \pm \frac{\sqrt{5}}{3} \end{aligned}$
$\begin{align*}9 \cos^2x-5&=0 \\ 9 \cos^2x&=5 \\ \cos^2x&=\frac{5}{9} \\ \cos x&=\pm \frac{\sqrt{5}}{3}\end{align*}$
::9cos2%x-5=09cos2x=5cos2x=59cos*x=59cos*x53

The $\begin{align*}\cos x=\frac{\sqrt{5}}{3}\end{align*}$ at $\begin{align*}x=0.243\end{align*}$ rad (use your graphing calculator). To find the other value where cosine is positive, subtract 0.243 from $\begin{align*}2 \pi\end{align*}$ , $\begin{align*}x=2 \pi -0.243=6.037\end{align*}$ rad.
::x=0.243 rad 的 cosx=53 (使用您的图形计算计算器) 。要找到其他正余弦值, 请从 2 x= 2 = 2. 0. 243 = 6. 037 rad 中减去 0. 243 。

The $\begin{align*}\cos x=- \frac{\sqrt{5}}{3}\end{align*}$ at $\begin{align*}x=2.412\end{align*}$ rad, which is in the $\begin{align*}2^{nd}\end{align*}$ quadrant. To find the other value where cosine is negative (the $\begin{align*}3^{rd}\end{align*}$ quadrant), use the reference angle , 0.243, and add it to $\begin{align*}\pi\end{align*}$ . $\begin{align*}x= \pi+0.243=3.383\end{align*}$ rad.
::x=2.412 rad 的 cosx53, 位于第二象限。要找到余弦为负( 第三象限)的其他值, 请使用 0. 243 的引用角度, 并将其添加到 x. x0. 243 =3.383 rad 。

Example 4
::例4

Solve the following trig equation in the interval $\begin{align*}0 \le x< 2 \pi\end{align*}$ .
::在 0. x < 2 间距内解决以下三重方程。

$\begin{align*}3 \sec(x-1)+2=0\end{align*}$
::3sec(x- 1)+2=0

Here, we will find the solution within the given range, $\begin{align*}0 \le x< 2 \pi\end{align*}$ .
::在这里,我们将在给定范围内找到解决方案, 0x<2。

$\begin{aligned} 3 \sec (x - 1) + 2 & = 0 \\ 3 \sec (x - 1) & = - 2 \\ \sec (x - 1) & = - \frac{2}{3} \\ \cos (x - 1) & = - \frac{3}{2} \end{aligned}$
$\begin{align*}3 \sec(x-1)+2&=0 \\ 3 \sec(x-1)&=-2 \\ \sec(x-1)&=- \frac{2}{3} \\ \cos(x-1)&=- \frac{3}{2}\end{align*}$
::3sec(x- 1) +2=03sec(x-1) 2sec(x-1) 2sec(x-1) 23cos(x-1) 32

At this point, we can stop. The range of the cosine function is from 1 to -1. $\begin{align*}- \frac{3}{2}\end{align*}$ is outside of this range, so there is no solution to this equation.
::此时,我们可以停止。余弦函数的范围从 1 到 - 1. 。 - 32 不属于此范围, 因此无法解决此等式。

Review
::回顾

Determine if the following values for $\begin{align*}x\end{align*}$ . are solutions to the equation $\begin{align*}5+6 \csc x=17\end{align*}$ .
::确定 x. 的以下值是否是公式 5+6cscx=17的解决方案。

$\begin{align*}x=- \frac{7 \pi}{6}\end{align*}$
::x76

$\begin{align*}x=\frac{11 \pi}{6}\end{align*}$
::x=116

$\begin{align*}x=\frac{5 \pi}{6}\end{align*}$
::x=56

Solve the following trigonometric equations. If no solutions exist, write no solution .
::解决以下三角方程式。如果没有解决方案, 请不要写入解决方案。

$\begin{align*}1- \cos x=0\end{align*}$
::1 - COsx=0

$\begin{align*}3 \tan x - \sqrt{3}=0\end{align*}$
::3tanx-3=0

$\begin{align*}4 \cos x=2 \cos x+1\end{align*}$
::4cosx=2cosx+1

$\begin{align*}5 \sin x-2=2 \sin x+4\end{align*}$
::5sinx-2=2sinx+4

$\begin{align*}\sec x-4=- \sec x\end{align*}$
::秒 *%x - 4sec*x

$\begin{align*}\tan^2(x-2)=3\end{align*}$
::tan2(x-2)=3

Sole the following trigonometric equations within the interval $\begin{align*}0 \le x < 2 \pi\end{align*}$ . If no solutions exist, write no solution .
::在 0. x < 2 的间距内将以下三角方程分离出来。如果没有解决方案, 请不要写入解决方案。

$\begin{align*}\cos x=\sin x\end{align*}$
::COsx=sinx

$\begin{align*}- \sqrt{3} \csc x=2\end{align*}$
::- 3cscx=2

$\begin{align*}6 \sin(x-2)=14\end{align*}$
::6sin(x-2)=14

$\begin{align*}7 \cos x -4=1\end{align*}$
::7cosx-4=1

$\begin{align*}5+4 \cot^2x=17\end{align*}$
::5+4cot2x=17

$\begin{align*}2 \sin^2x-7=-6\end{align*}$
::2 辛2 x-7 6

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

使用代数解决三角等量

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)