3.5 使用二次曲线公式的三角等量

章节大纲

• 选择节常规

  折叠 展开

  常规

  全部折叠 展开全部

  • 选择活动新闻通告

    

    新闻通告 讨论区
• 选择节使用二次曲线公式的三角等量公式

  折叠 展开

  使用二次曲线公式的三角等量公式

  播放段落

  Solving equations is a fundamental part of mathematics. Being able to find which values of a variable fit an equation allows us to determine all sorts of interesting behavior, both in math and in the sciences. Solving trig equations for angles that satisfy the equation is one application of mathematical methods for solving equations. Suppose someone gave you the following equation:
  ::解析方程式是数学的一个基本部分。 能够找到一个变量的哪个值适合一个方程式, 使我们能够在数学和科学中确定各种有趣的行为。 解析满足方程式角度的三角方程式是数学方法解析方程式的一个应用。 假设有人给了你以下方程式:

  播放段落

  $\begin{array}{r}3{\sin }^2\theta +8\sin \theta -3=0\end{array}$
  ::3辛28辛3=0

  播放段落

  Quadratic Functions with Trigonometric Equations
  ::具有三角等量的二次函数

  播放段落

  When solving quadratic equations that do not factor, the quadratic formula is often used.
  ::当解决不考虑的二次方程时,往往使用二次方程。

  播放段落

  Remember that the quadratic equation is:
  ::记住,二次方程是:

  播放段落

  $\begin{array}{r}a{x}^2+bx+c=0\end{array}$ (where a, b, and c are constants)
  ::ax2+bx+c=0(a、b和c为常数)

  播放段落

  In this situation, you can use the quadratic formula to find out what values of "x" satisfy the equation.
  ::在此情况下,您可以使用二次公式来找出“x”的值满足方程式。

  播放段落

  The same method can be applied when that do not factor. The values for $\begin{array}{r}a\end{array}$ is the numerical coefficient of the function's squared term, $\begin{array}{r}b\end{array}$ is the numerical coefficient of the function term that is to the first power and $\begin{array}{r}c\end{array}$ is a constant. The formula will result in two answers and both will have to be evaluated within the designated interval.
  ::当该方法不包含系数时,也可以使用同样的方法。a 的数值是函数方形术语的数字系数,b 函数术语的数值系数是第一个功率的数值系数,c 是一个常数。公式将产生两个答案,两者都必须在指定的间隔内评价。

  播放段落

  Solving for Unknown Values 
  ::解决未知值

  播放段落

  1. Solve $\begin{array}{r}3{\cot }^{2}x-3\cot x=1\end{array}$ for exact values of $\begin{array}{r}x\end{array}$ over the interval $\begin{array}{r}[0,2\pi ]\end{array}$ .
  ::1. 在间隔[0,2]内为x的准确值解决 3 个 cot2x-3 cotx=1 。

  播放段落

  $$\begin{array}{rl}3{\cot }^{2}x-3\cot x& =1\\ 3{\cot }^{2}x-3\cot x-1& =0\end{array}$$

  
  ::3cot2x-3cotx=13cot2x-3cotx-1=0

  播放段落

  The equation will not factor. Use the quadratic formula for $\begin{array}{r}\cot x\end{array}$ , $\begin{array}{r}a=3,b=-3,c=-1\end{array}$ .
  ::方程不会因数 。 使用 cotx 的二次公式, a= 3, b3, c1 。

  播放段落

  $$\begin{array}{rl}\cot x& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \cot x& =\frac{-(-3)\pm \sqrt{(-3{)}^2-4(3)(-1)}}{2(3)}\\ \cot x& =\frac{3\pm \sqrt{9+12}}{6}\\ \cot x& =\frac{3+\sqrt{21}}{6}\text{or}& & \cot x=\frac{3-\sqrt{21}}{6}\\ \cot x& =\frac{3+4.5826}{6}& & \cot x=\frac{3-4.5826}{6}\\ \cot x& =1.2638& & \cot x=-0.2638\\ \tan x& =\frac{1}{1.2638}& & \tan x=\frac{1}{-0.2638}\\ x& =0.6694,3.81099& & x=1.8287,4.9703\end{array}$$

  
  ::xxbb2-4ac2acotx(-3)(-3)(-3)(3)2-4(3)(3)(-1)、2(3)x=39+126cotx=3+216orcotx=3+216cotx=3+216cotx=3+4.58266cóx=3-4.58266cotx=1.2638cotxx1.2638cotx=11.2638tanx=10-0.2638x=0.6694,3.81099x=1.8287,4.9703。

  播放段落

  2. Solve $\begin{array}{r}-5{\cos }^{2}x+9\sin x+3=0\end{array}$ for values of $\begin{array}{r}x\end{array}$ over the interval $\begin{array}{r}[0,2\pi ]\end{array}$ .
  ::2. 在间隔[0,2]内为 x 的 x 值解决 - 5cos2+9sin%x+3=0 。

  播放段落

  Change $\begin{array}{r}{\cos }^{2}x\end{array}$ to $\begin{array}{r}1-{\sin }^{2}x\end{array}$ from the Pythagorean Identity .
  ::将 Pythagorean 身份更改为 Cos2_x 到 1- sin2_x 。

  播放段落

  $$\begin{array}{rl}-5{\cos }^{2}x+9\sin x+3& =0\\ -5(1-{\sin }^{2}x)+9\sin x+3& =0\\ -5+5{\sin }^{2}x+9\sin x+3& =0\\ 5{\sin }^{2}x+9\sin x-2& =0\end{array}$$

  
  ::-5cos2x+9sinx+3=0-5(1-sin2x)+9sinx+3=0-5+5sin2x+9sin*x+3=05sin2x+9sin_9sin*x+3=0)

  播放段落

  $$\begin{array}{rl}& \sin x=\frac{-9\pm \sqrt{9^2-4(5)(-2)}}{2(5)}\\ & \sin x=\frac{-9\pm \sqrt{81+40}}{10}\\ & \sin x=\frac{-9\pm \sqrt{121}}{10}\\ & \sin x=\frac{-9+11}{10}\text{and}\sin x=\frac{-9-11}{10}\\ & \sin x=\frac{1}{5}\text{and}-2\\ & {\sin }^{-1}(0.2)\text{and}{\sin }^{-1}(-2)\end{array}$$

  
  ::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

  播放段落

  $\begin{array}{r}x\approx .201rad\end{array}$ and $\begin{array}{r}\pi -.201\approx 2.941\end{array}$
  ::x.201 rad 和 .201 2.941

  播放段落

  These are the only solutions for $\begin{array}{r}x\end{array}$ since $\begin{array}{r}-2\end{array}$ is not in the range of values.
  ::这是x的唯一解决办法,因为-2不在数值范围内。

  播放段落

  3. Solve $\begin{array}{r}3{\sin }^{2}x-6\sin x+2=0\end{array}$ for values of $\begin{array}{r}x\end{array}$ over the interval $\begin{array}{r}[0,2\pi ]\end{array}$ .
  ::3. 在间隔[0,2]内为xxxxx+2=0的值解决 3sin2x-6sinx+2=0。

  播放段落

  
  

  $$\begin{array}{rl}3{\sin }^{2}x-6\sin x+2& =0\end{array}$$

  
  ::3sin2x-6sinx+2=0

  播放段落

  $$\begin{array}{rl}& \sin x=\frac{6\pm \sqrt{(-6{)}^2-4(3)(2)}}{2(3)}\\ & \sin x=\frac{6\pm \sqrt{36-24}}{6}\\ & \sin x=\frac{6\pm \sqrt{12}}{6}\\ & \sin x=\frac{6+3.46}{10}\text{and}\sin x=\frac{6-3.46}{10}\\ & \sin x=.946\text{and}.254\\ & {\sin }^{-1}(0.946)\text{and}{\sin }^{-1}(0.254)\end{array}$$

  
  ::x=6(-6)2-(4)(3)(2)(2)(2)(2)(3)(-6)x=636-246sinx=6126sinx=6+3.4610和sinx=6-3.4610sinx=946和254sin-1(0.946)和sin-1__(0.254)

  播放段落

  $\begin{array}{r}x\approx 71.08deg\end{array}$ and $\begin{array}{r}\approx 14.71deg\end{array}$
  ::71.08 deg 和 14.71 deg

  播放段落

  Examples
  ::实例

  播放段落

  Example 1
  ::例1

  播放段落

  Earlier, you were asked to solve  the equation :
  ::早些时候,有人要求你解答这个方程式:

  播放段落

  $\begin{array}{r}3{\sin }^2\theta +8\sin \theta -3=0\end{array}$
  ::3辛28辛3=0

  播放段落

  Using the quadratic formula, with $\begin{array}{r}a=3,b=8,c=-3\end{array}$ , we get:
  ::使用四方形方程式,以a=3,b=8,c=3,我们得到:

  播放段落

  $\begin{array}{r}\sin \theta =\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-8\pm \sqrt{64-(4)(3)(-3)}}{6}=\frac{-8\pm \sqrt{100}}{6}=\frac{-8\pm 10}{6}=\frac{1}{3}or-3\end{array}$
  ::b2-4ac2a864-(4)(3)(3)(3)(3)(3)(3)-6810068106=13or-3

  播放段落

  The solution of -3 is ignored because sine can't take that value, however:
  ::-3的解决方案被忽略了 因为Sinr不能接受这个价值, 但是:

  播放段落

  $\begin{array}{r}{\sin }^{-1}\frac{1}{3}={19.471}^{\circ }\end{array}$
  ::13=19.471 *

  播放段落

  Example 2
  ::例2

  播放段落

  Solve $\begin{array}{r}{\sin }^{2}x-2\sin x-3=0\end{array}$ for $\begin{array}{r}x\end{array}$ over $\begin{array}{r}[0,\pi ]\end{array}$ .
  ::对 x 超过 [0] 的 x 解除 sin2x- 2sinx- 3=0 。

  播放段落

  You can factor this one like a quadratic.
  ::你可以把这个因素当成二次方形来考虑

  播放段落

  $$\begin{array}{rl}& {\sin }^{2}x-2\sin x-3=0\\ & (\sin x-3)(\sin x+1)=0\\ & \sin x-3=0& & \sin x+1=0\\ & \sin x=3\text{or}& & \sin x=-1\\ & x={\sin }^{-1}(3)& & x=\frac{3\pi }{2}\end{array}$$

  
  ::sin_x_3 (sin_x_3 (sin_x+1) = 0sin_x_3 = 0 sin_x_3 = 0 sin_x+1 = 0 sin_x= 0 sin_x_1 xsin_1x=sin_1}(3) x=3}2

  播放段落

  For this problem the only solution is $\begin{array}{r}\frac{3\pi }{2}\end{array}$ because sine cannot be $\begin{array}{r}3\end{array}$ (it is not in the range).
  ::对于这个问题,唯一的解决办法是3-12,因为Sin不能是3(它不在幅度之内)。

  播放段落

  Example 3
  ::例3

  播放段落

  Solve $\begin{array}{r}{\tan }^{2}x+\tan x-2=0\end{array}$ for values of $\begin{array}{r}x\end{array}$ over the interval $\begin{array}{r}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\end{array}$ .
  ::对于间隔内 x 值的 x 值, 解决 tan2 = 2 = 0 [2 +2] 。

  播放段落

  $\begin{array}{r}{\tan }^{2}x+\tan x-2=0\end{array}$
  ::tan2_ x+tan_ x-2=0

  播放段落

  $$\begin{array}{rl}\frac{-1\pm \sqrt{1^2-4(1)(-2)}}{2}& =\tan x\\ \frac{-1\pm \sqrt{1+8}}{2}& =\tan x\\ \frac{-1\pm 3}{2}& =\tan x\\ \tan x& =-2\text{or}1\end{array}$$

  
  ::-112-4(1)(-1-2)2=tanx-11-1+82=tanx-132=tanxnxx2or1

  播放段落

  $\begin{array}{r}\tan x=1\end{array}$ when $\begin{array}{r}x=\frac{\pi }{4}\end{array}$ , in the interval $\begin{array}{r}\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\end{array}$
  ::x4 的间隔 [2,2] 时 tanx=1

  播放段落

  $\begin{array}{r}\tan x=-2\end{array}$ when $\begin{array}{r}x=-1.107rad\end{array}$
  ::x1.107 rad 时 tanx%2

  播放段落

  Example 4
  ::例4

  播放段落

  Solve the trigonometric equation such that $\begin{array}{r}5{\cos }^2\theta -6\sin \theta =0\end{array}$ over the interval $\begin{array}{r}[0,2\pi ]\end{array}$ .
  ::解决三角方程 5cos26sin0 在间隔[0,2]。

  播放段落

  $\begin{array}{r}5{\cos }^2\theta -6\sin \theta =0\end{array}$ over the interval $\begin{array}{r}[0,2\pi ]\end{array}$ .
  ::间隔5cos26sin0[0,2]。

  播放段落

  $$\begin{array}{rl}5\left(1-{\sin }^{2}x\right)-6\sin x& =0\\ -5{\sin }^{2}x-6\sin x+5& =0\\ 5{\sin }^{2}x+6\sin x-5& =0\\ \frac{-6\pm \sqrt{6^2-4(5)(-5)}}{2(5)}& =\sin x\\ \frac{-6\pm \sqrt{36+100}}{10}& =\sin x\\ \frac{-6\pm \sqrt{136}}{10}& =\sin x\\ \frac{-6\pm 2\sqrt{34}}{10}& =\sin x\\ \frac{-3\pm \sqrt{34}}{5}& =\sin x\end{array}$$

  
  ::5(1-辛2x)-6sinx=0-5sin2x-6sinx+5=05sin2xx+6sinx-5=0-662-4(5)(-5)-5=0-662-4(5)(-5),2(5)=sinx-636+10010=sinx-6613610=sinx-6__13610=sinx-623410=sinx623410=sin_x33×345=sinx

  播放段落

  $\begin{array}{r}x={\sin }^{-1}\left(\frac{-3+\sqrt{34}}{5}\right)\end{array}$ or $\begin{array}{r}{\sin }^{-1}\left(\frac{-3-\sqrt{34}}{5}\right)\end{array}$ $\begin{array}{r}x=0.6018rad\end{array}$ or $\begin{array}{r}2.5398rad\end{array}$ from the first expression, the second expression will not yield any answers because it is out the range of sine.
  ::x=sin-1(- 3+345) 或sin- 1(- 3- 345)x=0. 6018 rad 或 2. 5398 rad 从第一个表达式开始,第二个表达式不会产生任何答案,因为它超出了正弦范围 。

  播放段落

  Review
  ::回顾

  播放段落

  Solve each equation using the quadratic formula.
  ::使用二次方程式解决每个方程式。

  播放段落
  1. $\begin{array}{r}3x^2+10x+2=0\end{array}$
     ::3x2+10x+2=0
  播放段落
  2. $\begin{array}{r}5x^2+10x+2=0\end{array}$
     ::5x2+10x+2=0
  播放段落
  3. $\begin{array}{r}2x^2+6x-5=0\end{array}$
     ::2x2+6x-5=0

  播放段落

  Use the quadratic formula to solve each quadratic equation over the interval $\begin{array}{r}[0,2\pi )\end{array}$ .
  ::使用二次方程解析间隔[0,2]中的每一二次方程。

  播放段落
  4. $\begin{array}{r}3{\cos }^2(x)+10\cos (x)+2=0\end{array}$
     ::3cos2(x)+10cos(x)+2=0
  播放段落
  5. $\begin{array}{r}5{\sin }^2(x)+10\sin (x)+2=0\end{array}$
     ::5sin2(x)+10sin(x)+2=0
  播放段落
  6. $\begin{array}{r}2{\sin }^2(x)+6\sin (x)-5=0\end{array}$
     ::2sin2(x)+6sin(x)-5=0
  播放段落
  7. $\begin{array}{r}6{\cos }^2(x)-5\cos (x)-21=0\end{array}$
     ::6cos2(x)-5cos(x)-21=0
  播放段落
  8. $\begin{array}{r}9{\tan }^2(x)-42\tan (x)+49=0\end{array}$
     ::9tan2(x)-42tan(x)+49=0
  播放段落
  9. $\begin{array}{r}{\sin }^2(x)+3\sin (x)=5\end{array}$
     :x)+3sin *(x)=5
  播放段落
  10. $\begin{array}{r}3{\cos }^2(x)-4\sin (x)=0\end{array}$
      ::3cos2(x)- 4sin(x)=0
  播放段落
  11. $\begin{array}{r}-2{\cos }^2(x)+4\sin (x)=0\end{array}$
      ::-2cos2(x)+4sin(x)=0
  播放段落
  12. $\begin{array}{r}{\tan }^2(x)+\tan (x)=3\end{array}$
      ::tan(x)=3
  播放段落
  13. $\begin{array}{r}{\cot }^2(x)+5\tan (x)+14=0\end{array}$
      ::comt2(x)+5tan(x)+14=0
  播放段落
  14. $\begin{array}{r}{\sin }^2(x)+\sin (x)=1\end{array}$
      ::sin2(x)+sin}(x)=1
  播放段落
  15. What type of sine or cosine equations have no solution?
      ::哪种正弦或共弦方程式没有解决办法?

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。