解答
20=27sin(x)−1.5cos(x)
解答
x=2.36461…+2πn,x=0.88797…+2πn
+1
度数
x=135.48245…∘+360∘n,x=50.87720…∘+360∘n求解步骤
20=27sin(x)−1.5cos(x)
两边加上 1.5cos(x)27sin(x)=20+1.5cos(x)
两边进行平方(27sin(x))2=(20+1.5cos(x))2
两边减去 (20+1.5cos(x))2729sin2(x)−400−60cos(x)−2.25cos2(x)=0
使用三角恒等式改写
−400−2.25cos2(x)−60cos(x)+729sin2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−400−2.25cos2(x)−60cos(x)+729(1−cos2(x))
化简 −400−2.25cos2(x)−60cos(x)+729(1−cos2(x)):−731.25cos2(x)−60cos(x)+329
−400−2.25cos2(x)−60cos(x)+729(1−cos2(x))
乘开 729(1−cos2(x)):729−729cos2(x)
729(1−cos2(x))
使用分配律: a(b−c)=ab−aca=729,b=1,c=cos2(x)=729⋅1−729cos2(x)
数字相乘:729⋅1=729=729−729cos2(x)
=−400−2.25cos2(x)−60cos(x)+729−729cos2(x)
化简 −400−2.25cos2(x)−60cos(x)+729−729cos2(x):−731.25cos2(x)−60cos(x)+329
−400−2.25cos2(x)−60cos(x)+729−729cos2(x)
对同类项分组=−2.25cos2(x)−60cos(x)−729cos2(x)−400+729
同类项相加:−2.25cos2(x)−729cos2(x)=−731.25cos2(x)=−731.25cos2(x)−60cos(x)−400+729
数字相加/相减:−400+729=329=−731.25cos2(x)−60cos(x)+329
=−731.25cos2(x)−60cos(x)+329
=−731.25cos2(x)−60cos(x)+329
329−60cos(x)−731.25cos2(x)=0
用替代法求解
329−60cos(x)−731.25cos2(x)=0
令:cos(x)=u329−60u−731.25u2=0
329−60u−731.25u2=0:u=−1462506000+9659250000,u=1462509659250000−6000
329−60u−731.25u2=0
在两边乘以 100
329−60u−731.25u2=0
To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are 2digits to the right of the decimal point, therefore multiply by 100329⋅100−60u⋅100−731.25u2⋅100=0⋅100
整理后得32900−6000u−73125u2=0
32900−6000u−73125u2=0
改写成标准形式 ax2+bx+c=0−73125u2−6000u+32900=0
使用求根公式求解
−73125u2−6000u+32900=0
二次方程求根公式:
若 a=−73125,b=−6000,c=32900u1,2=2(−73125)−(−6000)±(−6000)2−4(−73125)⋅32900
u1,2=2(−73125)−(−6000)±(−6000)2−4(−73125)⋅32900
(−6000)2−4(−73125)⋅32900=9659250000
(−6000)2−4(−73125)⋅32900
使用法则 −(−a)=a=(−6000)2+4⋅73125⋅32900
使用指数法则: (−a)n=an,若 n 是偶数(−6000)2=60002=60002+4⋅73125⋅32900
数字相乘:4⋅73125⋅32900=9623250000=60002+9623250000
60002=36000000=36000000+9623250000
数字相加:36000000+9623250000=9659250000=9659250000
u1,2=2(−73125)−(−6000)±9659250000
将解分隔开u1=2(−73125)−(−6000)+9659250000,u2=2(−73125)−(−6000)−9659250000
u=2(−73125)−(−6000)+9659250000:−1462506000+9659250000
2(−73125)−(−6000)+9659250000
去除括号: (−a)=−a,−(−a)=a=−2⋅731256000+9659250000
数字相乘:2⋅73125=146250=−1462506000+9659250000
使用分式法则: −ba=−ba=−1462506000+9659250000
u=2(−73125)−(−6000)−9659250000:1462509659250000−6000
2(−73125)−(−6000)−9659250000
去除括号: (−a)=−a,−(−a)=a=−2⋅731256000−9659250000
数字相乘:2⋅73125=146250=−1462506000−9659250000
使用分式法则: −b−a=ba6000−9659250000=−(9659250000−6000)=1462509659250000−6000
二次方程组的解是:u=−1462506000+9659250000,u=1462509659250000−6000
u=cos(x)代回cos(x)=−1462506000+9659250000,cos(x)=1462509659250000−6000
cos(x)=−1462506000+9659250000,cos(x)=1462509659250000−6000
cos(x)=−1462506000+9659250000:x=arccos(−1462506000+9659250000)+2πn,x=−arccos(−1462506000+9659250000)+2πn
cos(x)=−1462506000+9659250000
使用反三角函数性质
cos(x)=−1462506000+9659250000
cos(x)=−1462506000+9659250000的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−1462506000+9659250000)+2πn,x=−arccos(−1462506000+9659250000)+2πn
x=arccos(−1462506000+9659250000)+2πn,x=−arccos(−1462506000+9659250000)+2πn
cos(x)=1462509659250000−6000:x=arccos(1462509659250000−6000)+2πn,x=2π−arccos(1462509659250000−6000)+2πn
cos(x)=1462509659250000−6000
使用反三角函数性质
cos(x)=1462509659250000−6000
cos(x)=1462509659250000−6000的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(1462509659250000−6000)+2πn,x=2π−arccos(1462509659250000−6000)+2πn
x=arccos(1462509659250000−6000)+2πn,x=2π−arccos(1462509659250000−6000)+2πn
合并所有解x=arccos(−1462506000+9659250000)+2πn,x=−arccos(−1462506000+9659250000)+2πn,x=arccos(1462509659250000−6000)+2πn,x=2π−arccos(1462509659250000−6000)+2πn
将解代入原方程进行验证
将它们代入 27sin(x)−1.5cos(x)=20检验解是否符合
去除与方程不符的解。
检验 arccos(−1462506000+9659250000)+2πn的解:真
arccos(−1462506000+9659250000)+2πn
代入 n=1arccos(−1462506000+9659250000)+2π1
对于 27sin(x)−1.5cos(x)=20代入x=arccos(−1462506000+9659250000)+2π127sin(arccos(−1462506000+9659250000)+2π1)−1.5cos(arccos(−1462506000+9659250000)+2π1)=20
整理后得20=20
⇒真
检验 −arccos(−1462506000+9659250000)+2πn的解:假
−arccos(−1462506000+9659250000)+2πn
代入 n=1−arccos(−1462506000+9659250000)+2π1
对于 27sin(x)−1.5cos(x)=20代入x=−arccos(−1462506000+9659250000)+2π127sin(−arccos(−1462506000+9659250000)+2π1)−1.5cos(−arccos(−1462506000+9659250000)+2π1)=20
整理后得−17.86089…=20
⇒假
检验 arccos(1462509659250000−6000)+2πn的解:真
arccos(1462509659250000−6000)+2πn
代入 n=1arccos(1462509659250000−6000)+2π1
对于 27sin(x)−1.5cos(x)=20代入x=arccos(1462509659250000−6000)+2π127sin(arccos(1462509659250000−6000)+2π1)−1.5cos(arccos(1462509659250000−6000)+2π1)=20
整理后得20=20
⇒真
检验 2π−arccos(1462509659250000−6000)+2πn的解:假
2π−arccos(1462509659250000−6000)+2πn
代入 n=12π−arccos(1462509659250000−6000)+2π1
对于 27sin(x)−1.5cos(x)=20代入x=2π−arccos(1462509659250000−6000)+2π127sin(2π−arccos(1462509659250000−6000)+2π1)−1.5cos(2π−arccos(1462509659250000−6000)+2π1)=20
整理后得−21.89295…=20
⇒假
x=arccos(−1462506000+9659250000)+2πn,x=arccos(1462509659250000−6000)+2πn
以小数形式表示解x=2.36461…+2πn,x=0.88797…+2πn