解答
73500=130000⋅sin(x)+0.15⋅130000⋅cos(x)
解答
x=2.39936…+2πn,x=0.44444…+2πn
+1
度数
x=137.47362…∘+360∘n,x=25.46484…∘+360∘n求解步骤
73500=130000sin(x)+0.15⋅130000cos(x)
两边减去 0.15130000cos(x)130000sin(x)=73500−19500cos(x)
两边进行平方(130000sin(x))2=(73500−19500cos(x))2
两边减去 (73500−19500cos(x))21300002sin2(x)−735002+2866500000cos(x)−380250000cos2(x)=0
使用三角恒等式改写
−735002+1300002sin2(x)+2866500000cos(x)−380250000cos2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−735002+1300002(1−cos2(x))+2866500000cos(x)−380250000cos2(x)
−735002+(1−cos2(x))⋅1300002+2866500000cos(x)−380250000cos2(x)=0
用替代法求解
−735002+(1−cos2(x))⋅1300002+2866500000cos(x)−380250000cos2(x)=0
令:cos(x)=u−735002+(1−u2)⋅1300002+2866500000u−380250000u2=0
−735002+(1−u2)⋅1300002+2866500000u−380250000u2=0:u=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000,u=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
−735002+(1−u2)⋅1300002+2866500000u−380250000u2=0
展开 −735002+(1−u2)⋅1300002+2866500000u−380250000u2:−735002+1300002−1300002u2+2866500000u−380250000u2
−735002+(1−u2)⋅1300002+2866500000u−380250000u2
=−735002+1300002(1−u2)+2866500000u−380250000u2
乘开 1300002(1−u2):1300002−1300002u2
1300002(1−u2)
使用分配律: a(b−c)=ab−aca=1300002,b=1,c=u2=1300002⋅1−1300002u2
乘以:1300002⋅1=1300002=1300002−1300002u2
=−735002+1300002−1300002u2+2866500000u−380250000u2
−735002+1300002−1300002u2+2866500000u−380250000u2=0
改写成标准形式 ax2+bx+c=0−(1300002+380250000)u2+2866500000u−735002+1300002=0
使用求根公式求解
−(1300002+380250000)u2+2866500000u−735002+1300002=0
二次方程求根公式:
若 a=−1300002−380250000,b=2866500000,c=−735002+1300002u1,2=2(−1300002−380250000)−2866500000±28665000002−4(−1300002−380250000)(−735002+1300002)
u1,2=2(−1300002−380250000)−2866500000±28665000002−4(−1300002−380250000)(−735002+1300002)
28665000002−4(−1300002−380250000)(−735002+1300002)=28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
28665000002−4(−1300002−380250000)(−735002+1300002)
乘开 28665000002−4(−1300002−380250000)(−735002+1300002):28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
28665000002−4(−1300002−380250000)(−735002+1300002)
乘开 −4(−1300002−380250000)(−735002+1300002):−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
乘开 (−1300002−380250000)(−735002+1300002):514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000
(−1300002−380250000)(−735002+1300002)
使用 FOIL 方法: (a+b)(c+d)=ac+ad+bc+bda=−1300002,b=−380250000,c=−735002,d=1300002=(−1300002)(−735002)+(−1300002)⋅1300002+(−380250000)(−735002)+(−380250000)⋅1300002
使用加减运算法则(−a)(−b)=ab,+(−a)=−a=1300002⋅735002−1300002⋅1300002+735002⋅380250000−1300002⋅380250000
化简 1300002⋅735002−1300002⋅1300002+735002⋅380250000−1300002⋅380250000:514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000
1300002⋅735002−1300002⋅1300002+735002⋅380250000−1300002⋅380250000
1300002⋅735002=514⋅14958268416
1300002⋅735002
分解整数 130000=24⋅54⋅13=(24⋅54⋅13)2⋅735002
使用指数法则: (ab)c=acbc(24⋅54⋅13)2=(24)2(54)2⋅132=(24)2(54)2⋅132⋅735002
使用指数法则: (ab)c=abc(24)2=24⋅2,(54)2=54⋅2=24⋅2⋅54⋅2⋅132⋅735002
整理后得=28⋅58⋅132⋅735002
分解整数 73500=53⋅22⋅147=28⋅58⋅132(22⋅53⋅147)2
使用指数法则: (ab)c=acbc(22⋅53⋅147)2=(22)2(53)2⋅1472=28⋅58⋅132(22)2(53)2⋅1472
使用指数法则: (ab)c=abc(22)2=22⋅2,(53)2=53⋅2=28⋅58⋅132⋅22⋅2⋅53⋅2⋅1472
整理后得=28⋅58⋅132⋅24⋅56⋅1472
使用指数法则: ab⋅ac=ab+c28⋅24=28+4=58⋅132⋅28+4⋅56⋅1472
数字相加:8+4=12=58⋅132⋅212⋅56⋅1472
使用指数法则: ab⋅ac=ab+c58⋅56=58+6=132⋅212⋅58+6⋅1472
数字相加:8+6=14=132⋅212⋅514⋅1472
132=169=514⋅212⋅1472⋅169
212=4096=514⋅1472⋅169⋅4096
1472=21609=514⋅169⋅4096⋅21609
数字相乘:169⋅4096⋅21609=14958268416=514⋅14958268416
1300002⋅1300002=1300004
1300002⋅1300002
使用指数法则: ab⋅ac=ab+c1300002⋅1300002=1300002+2=1300002+2
数字相加:2+2=4=1300004
=514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000
=514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000
=−4(514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000)
乘开 −4(514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000):−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
−4(514⋅14958268416−1300004+735002⋅380250000−1300002⋅380250000)
打开括号=(−4)⋅514⋅14958268416+(−4)(−1300004)+(−4)⋅735002⋅380250000+(−4)(−1300002⋅380250000)
使用加减运算法则+(−a)=−a,(−a)(−b)=ab=−514⋅4⋅14958268416+1300004⋅4−735002⋅4⋅380250000+1300002⋅4⋅380250000
化简 −514⋅4⋅14958268416+1300004⋅4−735002⋅4⋅380250000+1300002⋅4⋅380250000:−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
−514⋅4⋅14958268416+1300004⋅4−735002⋅4⋅380250000+1300002⋅4⋅380250000
数字相乘:4⋅14958268416=59833073664=−514⋅59833073664+1300004⋅4−735002⋅4⋅380250000+1300002⋅4⋅380250000
数字相乘:4⋅380250000=1521000000=−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
=−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
=−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
=28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
=28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
u1,2=2(−1300002−380250000)−2866500000±28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
将解分隔开u1=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000,u2=2(−1300002−380250000)−2866500000−28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
u=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
u=2(−1300002−380250000)−2866500000−28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000:−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
2(−1300002−380250000)−2866500000−28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
使用分式法则: −b−a=ba−2866500000−28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000=−(1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
使用分式法则: −ba=−ba=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
二次方程组的解是:u=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000,u=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
u=cos(x)代回cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000,cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000,cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000:x=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn
cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
使用反三角函数性质
cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000
cos(x)=2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn
x=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn
cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000:x=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn,x=2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
使用反三角函数性质
cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000
cos(x)=−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn,x=2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
x=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn,x=2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
合并所有解x=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn,x=2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
将解代入原方程进行验证
将它们代入 130000sin(x)+0.15130000cos(x)=73500检验解是否符合
去除与方程不符的解。
检验 arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn的解:真
arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn
代入 n=1arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1
对于 130000sin(x)+0.15130000cos(x)=73500代入x=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1130000sin(arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1)+0.15⋅130000cos(arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1)=73500
整理后得73500=73500
⇒真
检验 −arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn的解:假
−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn
代入 n=1−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1
对于 130000sin(x)+0.15130000cos(x)=73500代入x=−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1130000sin(−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1)+0.15⋅130000cos(−arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2π1)=73500
整理后得−102241.68342…=73500
⇒假
检验 arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn的解:真
arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
代入 n=1arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1
对于 130000sin(x)+0.15130000cos(x)=73500代入x=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1130000sin(arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1)+0.15⋅130000cos(arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1)=73500
整理后得73500=73500
⇒真
检验 2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn的解:假
2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
代入 n=12π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1
对于 130000sin(x)+0.15130000cos(x)=73500代入x=2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1130000sin(2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1)+0.15⋅130000cos(2π−arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2π1)=73500
整理后得−38288.87892…=73500
⇒假
x=arccos(2(−1300002−380250000)−2866500000+28665000002−514⋅59833073664+1300004⋅4−735002⋅1521000000+1300002⋅1521000000)+2πn,x=arccos(−2(−1300002−380250000)1300004⋅4+28665000002+1300002⋅1521000000−514⋅59833073664−735002⋅1521000000+2866500000)+2πn
以小数形式表示解x=2.39936…+2πn,x=0.44444…+2πn