解题
积分(反导数)计算器导数计算器代数计算器矩阵计算器更多的...
图表
线图指数图二次图正弦图更多的...
计算器
体质指数计算器复利计算器百分比计算器加速度计算器更多的...
几何
勾股定理计算器圆形面积计算器等腰三角形计算器三角形计算器更多的...
AI Chat
工具
笔记簿小组主题工作表练习验证
zs
English
Español
Português
Français
Deutsch
Italiano
Русский
中文(简体)
한국어
日本語
Tiếng Việt
עברית
العربية
受欢迎的 三角函数 >

21.16=19.6sin(x)-29.4cos(x)

  • 初等代数
  • 代数
  • 微积分入门
  • 微积分
  • 函数
  • 线性代数
  • 三角
  • 统计
  • 化学

解答

21.16=19.6sin(x)−29.4cos(x)

解答

x=−2.80086…+2πn,x=1.62485…+2πn
+1
度数
x=−160.47763…∘+360∘n,x=93.09749…∘+360∘n
求解步骤
21.16=19.6sin(x)−29.4cos(x)
两边加上 29.4cos(x)19.6sin(x)=21.16+29.4cos(x)
两边进行平方(19.6sin(x))2=(21.16+29.4cos(x))2
两边减去 (21.16+29.4cos(x))2384.16sin2(x)−447.7456−1244.208cos(x)−864.36cos2(x)=0
使用三角恒等式改写
−447.7456−1244.208cos(x)+384.16sin2(x)−864.36cos2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−447.7456−1244.208cos(x)+384.16(1−cos2(x))−864.36cos2(x)
化简 −447.7456−1244.208cos(x)+384.16(1−cos2(x))−864.36cos2(x):−1248.52cos2(x)−1244.208cos(x)−63.5856
−447.7456−1244.208cos(x)+384.16(1−cos2(x))−864.36cos2(x)
乘开 384.16(1−cos2(x)):384.16−384.16cos2(x)
384.16(1−cos2(x))
使用分配律: a(b−c)=ab−aca=384.16,b=1,c=cos2(x)=384.16⋅1−384.16cos2(x)
=1⋅384.16−384.16cos2(x)
数字相乘:1⋅384.16=384.16=384.16−384.16cos2(x)
=−447.7456−1244.208cos(x)+384.16−384.16cos2(x)−864.36cos2(x)
化简 −447.7456−1244.208cos(x)+384.16−384.16cos2(x)−864.36cos2(x):−1248.52cos2(x)−1244.208cos(x)−63.5856
−447.7456−1244.208cos(x)+384.16−384.16cos2(x)−864.36cos2(x)
同类项相加:−384.16cos2(x)−864.36cos2(x)=−1248.52cos2(x)=−447.7456−1244.208cos(x)+384.16−1248.52cos2(x)
对同类项分组=−1244.208cos(x)−1248.52cos2(x)−447.7456+384.16
数字相加/相减:−447.7456+384.16=−63.5856=−1248.52cos2(x)−1244.208cos(x)−63.5856
=−1248.52cos2(x)−1244.208cos(x)−63.5856
=−1248.52cos2(x)−1244.208cos(x)−63.5856
−63.5856−1244.208cos(x)−1248.52cos2(x)=0
用替代法求解
−63.5856−1244.208cos(x)−1248.52cos2(x)=0
令:cos(x)=u−63.5856−1244.208u−1248.52u2=0
−63.5856−1244.208u−1248.52u2=0:u=−2497.041244.208+1230501.974016​​,u=−2497.041244.208−1230501.974016​​
−63.5856−1244.208u−1248.52u2=0
改写成标准形式 ax2+bx+c=0−1248.52u2−1244.208u−63.5856=0
使用求根公式求解
−1248.52u2−1244.208u−63.5856=0
二次方程求根公式:
若 a=−1248.52,b=−1244.208,c=−63.5856u1,2​=2(−1248.52)−(−1244.208)±(−1244.208)2−4(−1248.52)(−63.5856)​​
u1,2​=2(−1248.52)−(−1244.208)±(−1244.208)2−4(−1248.52)(−63.5856)​​
(−1244.208)2−4(−1248.52)(−63.5856)​=1230501.974016​
(−1244.208)2−4(−1248.52)(−63.5856)​
使用法则 −(−a)=a=(−1244.208)2−4⋅1248.52⋅63.5856​
使用指数法则: (−a)n=an,若 n 是偶数(−1244.208)2=1244.2082=1244.2082−4⋅63.5856⋅1248.52​
数字相乘:4⋅1248.52⋅63.5856=317551.573248=1244.2082−317551.573248​
1244.2082=1548053.547264=1548053.547264−317551.573248​
数字相减:1548053.547264−317551.573248=1230501.974016=1230501.974016​
u1,2​=2(−1248.52)−(−1244.208)±1230501.974016​​
将解分隔开u1​=2(−1248.52)−(−1244.208)+1230501.974016​​,u2​=2(−1248.52)−(−1244.208)−1230501.974016​​
u=2(−1248.52)−(−1244.208)+1230501.974016​​:−2497.041244.208+1230501.974016​​
2(−1248.52)−(−1244.208)+1230501.974016​​
去除括号: (−a)=−a,−(−a)=a=−2⋅1248.521244.208+1230501.974016​​
数字相乘:2⋅1248.52=2497.04=−2497.041244.208+1230501.974016​​
使用分式法则: −ba​=−ba​=−2497.041244.208+1230501.974016​​
u=2(−1248.52)−(−1244.208)−1230501.974016​​:−2497.041244.208−1230501.974016​​
2(−1248.52)−(−1244.208)−1230501.974016​​
去除括号: (−a)=−a,−(−a)=a=−2⋅1248.521244.208−1230501.974016​​
数字相乘:2⋅1248.52=2497.04=−2497.041244.208−1230501.974016​​
使用分式法则: −ba​=−ba​=−2497.041244.208−1230501.974016​​
二次方程组的解是:u=−2497.041244.208+1230501.974016​​,u=−2497.041244.208−1230501.974016​​
u=cos(x)代回cos(x)=−2497.041244.208+1230501.974016​​,cos(x)=−2497.041244.208−1230501.974016​​
cos(x)=−2497.041244.208+1230501.974016​​,cos(x)=−2497.041244.208−1230501.974016​​
cos(x)=−2497.041244.208+1230501.974016​​:x=arccos(−2497.041244.208+1230501.974016​​)+2πn,x=−arccos(−2497.041244.208+1230501.974016​​)+2πn
cos(x)=−2497.041244.208+1230501.974016​​
使用反三角函数性质
cos(x)=−2497.041244.208+1230501.974016​​
cos(x)=−2497.041244.208+1230501.974016​​的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−2497.041244.208+1230501.974016​​)+2πn,x=−arccos(−2497.041244.208+1230501.974016​​)+2πn
x=arccos(−2497.041244.208+1230501.974016​​)+2πn,x=−arccos(−2497.041244.208+1230501.974016​​)+2πn
cos(x)=−2497.041244.208−1230501.974016​​:x=arccos(−2497.041244.208−1230501.974016​​)+2πn,x=−arccos(−2497.041244.208−1230501.974016​​)+2πn
cos(x)=−2497.041244.208−1230501.974016​​
使用反三角函数性质
cos(x)=−2497.041244.208−1230501.974016​​
cos(x)=−2497.041244.208−1230501.974016​​的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−2497.041244.208−1230501.974016​​)+2πn,x=−arccos(−2497.041244.208−1230501.974016​​)+2πn
x=arccos(−2497.041244.208−1230501.974016​​)+2πn,x=−arccos(−2497.041244.208−1230501.974016​​)+2πn
合并所有解x=arccos(−2497.041244.208+1230501.974016​​)+2πn,x=−arccos(−2497.041244.208+1230501.974016​​)+2πn,x=arccos(−2497.041244.208−1230501.974016​​)+2πn,x=−arccos(−2497.041244.208−1230501.974016​​)+2πn
将解代入原方程进行验证
将它们代入 19.6sin(x)−29.4cos(x)=21.16检验解是否符合
去除与方程不符的解。
检验 arccos(−2497.041244.208+1230501.974016​​)+2πn的解:假
arccos(−2497.041244.208+1230501.974016​​)+2πn
代入 n=1arccos(−2497.041244.208+1230501.974016​​)+2π1
对于 19.6sin(x)−29.4cos(x)=21.16代入x=arccos(−2497.041244.208+1230501.974016​​)+2π119.6sin(arccos(−2497.041244.208+1230501.974016​​)+2π1)−29.4cos(arccos(−2497.041244.208+1230501.974016​​)+2π1)=21.16
整理后得34.25965…=21.16
⇒假
检验 −arccos(−2497.041244.208+1230501.974016​​)+2πn的解:真
−arccos(−2497.041244.208+1230501.974016​​)+2πn
代入 n=1−arccos(−2497.041244.208+1230501.974016​​)+2π1
对于 19.6sin(x)−29.4cos(x)=21.16代入x=−arccos(−2497.041244.208+1230501.974016​​)+2π119.6sin(−arccos(−2497.041244.208+1230501.974016​​)+2π1)−29.4cos(−arccos(−2497.041244.208+1230501.974016​​)+2π1)=21.16
整理后得21.16=21.16
⇒真
检验 arccos(−2497.041244.208−1230501.974016​​)+2πn的解:真
arccos(−2497.041244.208−1230501.974016​​)+2πn
代入 n=1arccos(−2497.041244.208−1230501.974016​​)+2π1
对于 19.6sin(x)−29.4cos(x)=21.16代入x=arccos(−2497.041244.208−1230501.974016​​)+2π119.6sin(arccos(−2497.041244.208−1230501.974016​​)+2π1)−29.4cos(arccos(−2497.041244.208−1230501.974016​​)+2π1)=21.16
整理后得21.16=21.16
⇒真
检验 −arccos(−2497.041244.208−1230501.974016​​)+2πn的解:假
−arccos(−2497.041244.208−1230501.974016​​)+2πn
代入 n=1−arccos(−2497.041244.208−1230501.974016​​)+2π1
对于 19.6sin(x)−29.4cos(x)=21.16代入x=−arccos(−2497.041244.208−1230501.974016​​)+2π119.6sin(−arccos(−2497.041244.208−1230501.974016​​)+2π1)−29.4cos(−arccos(−2497.041244.208−1230501.974016​​)+2π1)=21.16
整理后得−17.98273…=21.16
⇒假
x=−arccos(−2497.041244.208+1230501.974016​​)+2πn,x=arccos(−2497.041244.208−1230501.974016​​)+2πn
以小数形式表示解x=−2.80086…+2πn,x=1.62485…+2πn

作图

Sorry, your browser does not support this application
查看交互式图形

流行的例子

10sin(x)cos^2(x)+3cos(x)=010sin(x)cos2(x)+3cos(x)=0(2cos(x))-1=0(2cos(x))−1=03cos(x)=sin^2(x)3cos(x)=sin2(x)sin(θ)=(5pi)/3sin(θ)=35π​10cos(θ)=810cos(θ)=8
学习工具人工智能数学求解器AI Chat工作表练习主题计算器作图计算器几何计算器验证解决方案
应用Symbolab 应用程序 (Android)作图计算器 (Android)练习 (Android)Symbolab 应用程序 (iOS)作图计算器 (iOS)练习 (iOS)Chrome 扩展程序
公司关于 Symbolab日志帮助
合法的隐私权Service TermsCookie 政策Cookie 设置请勿出售或分享我的个人信息版权、社区准则、DSA 和其他法律资源Learneo 法律中心
社交媒体
Symbolab, a Learneo, Inc. business
© Learneo, Inc. 2024