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受欢迎的 三角函数 >

sin(3x)=2sin(x)

  • 初等代数
  • 代数
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解答

sin(3x)=2sin(x)

解答

x=2πn,x=π+2πn,x=67π​+2πn,x=611π​+2πn,x=6π​+2πn,x=65π​+2πn
+1
度数
x=0∘+360∘n,x=180∘+360∘n,x=210∘+360∘n,x=330∘+360∘n,x=30∘+360∘n,x=150∘+360∘n
求解步骤
sin(3x)=2sin(x)
两边减去 2sin(x)sin(3x)−2sin(x)=0
使用三角恒等式改写
sin(3x)−2sin(x)
sin(3x)=3sin(x)−4sin3(x)
sin(3x)
使用三角恒等式改写
sin(3x)
改写为=sin(2x+x)
使用角和恒等式: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=sin(2x)cos(x)+cos(2x)sin(x)
使用倍角公式: sin(2x)=2sin(x)cos(x)=cos(2x)sin(x)+cos(x)2sin(x)cos(x)
化简 cos(2x)sin(x)+cos(x)⋅2sin(x)cos(x):sin(x)cos(2x)+2cos2(x)sin(x)
cos(2x)sin(x)+cos(x)2sin(x)cos(x)
cos(x)⋅2sin(x)cos(x)=2cos2(x)sin(x)
cos(x)2sin(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=2sin(x)cos1+1(x)
数字相加:1+1=2=2sin(x)cos2(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
=sin(x)cos(2x)+2cos2(x)sin(x)
使用倍角公式: cos(2x)=1−2sin2(x)=(1−2sin2(x))sin(x)+2cos2(x)sin(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
乘开 (1−2sin2(x))sin(x)+2(1−sin2(x))sin(x):−4sin3(x)+3sin(x)
(1−2sin2(x))sin(x)+2(1−sin2(x))sin(x)
=sin(x)(1−2sin2(x))+2sin(x)(1−sin2(x))
乘开 sin(x)(1−2sin2(x)):sin(x)−2sin3(x)
sin(x)(1−2sin2(x))
使用分配律: a(b−c)=ab−aca=sin(x),b=1,c=2sin2(x)=sin(x)1−sin(x)2sin2(x)
=1sin(x)−2sin2(x)sin(x)
化简 1⋅sin(x)−2sin2(x)sin(x):sin(x)−2sin3(x)
1sin(x)−2sin2(x)sin(x)
1⋅sin(x)=sin(x)
1sin(x)
乘以:1⋅sin(x)=sin(x)=sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
使用指数法则: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
数字相加:2+1=3=2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2(1−sin2(x))sin(x)
乘开 2sin(x)(1−sin2(x)):2sin(x)−2sin3(x)
2sin(x)(1−sin2(x))
使用分配律: a(b−c)=ab−aca=2sin(x),b=1,c=sin2(x)=2sin(x)1−2sin(x)sin2(x)
=2⋅1sin(x)−2sin2(x)sin(x)
化简 2⋅1⋅sin(x)−2sin2(x)sin(x):2sin(x)−2sin3(x)
2⋅1sin(x)−2sin2(x)sin(x)
2⋅1⋅sin(x)=2sin(x)
2⋅1sin(x)
数字相乘:2⋅1=2=2sin(x)
2sin2(x)sin(x)=2sin3(x)
2sin2(x)sin(x)
使用指数法则: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=2sin2+1(x)
数字相加:2+1=3=2sin3(x)
=2sin(x)−2sin3(x)
=2sin(x)−2sin3(x)
=sin(x)−2sin3(x)+2sin(x)−2sin3(x)
化简 sin(x)−2sin3(x)+2sin(x)−2sin3(x):−4sin3(x)+3sin(x)
sin(x)−2sin3(x)+2sin(x)−2sin3(x)
对同类项分组=−2sin3(x)−2sin3(x)+sin(x)+2sin(x)
同类项相加:−2sin3(x)−2sin3(x)=−4sin3(x)=−4sin3(x)+sin(x)+2sin(x)
同类项相加:sin(x)+2sin(x)=3sin(x)=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=−4sin3(x)+3sin(x)
=3sin(x)−4sin3(x)−2sin(x)
化简=sin(x)−4sin3(x)
sin(x)−4sin3(x)=0
用替代法求解
sin(x)−4sin3(x)=0
令:sin(x)=uu−4u3=0
u−4u3=0:u=0,u=−21​,u=21​
u−4u3=0
因式分解 u−4u3:−u(2u+1)(2u−1)
u−4u3
因式分解出通项 −u:−u(4u2−1)
−4u3+u
使用指数法则: ab+c=abacu3=u2u=−4u2u+u
因式分解出通项 −u=−u(4u2−1)
=−u(4u2−1)
分解 4u2−1:(2u+1)(2u−1)
4u2−1
将 4u2−1 改写为 (2u)2−12
4u2−1
将 4 改写为 22=22u2−1
将 1 改写为 12=22u2−12
使用指数法则: ambm=(ab)m22u2=(2u)2=(2u)2−12
=(2u)2−12
使用平方差公式: x2−y2=(x+y)(x−y)(2u)2−12=(2u+1)(2u−1)=(2u+1)(2u−1)
=−u(2u+1)(2u−1)
−u(2u+1)(2u−1)=0
使用零因数法则: If ab=0then a=0or b=0u=0or2u+1=0or2u−1=0
解 2u+1=0:u=−21​
2u+1=0
将 1到右边
2u+1=0
两边减去 12u+1−1=0−1
化简2u=−1
2u=−1
两边除以 2
2u=−1
两边除以 222u​=2−1​
化简u=−21​
u=−21​
解 2u−1=0:u=21​
2u−1=0
将 1到右边
2u−1=0
两边加上 12u−1+1=0+1
化简2u=1
2u=1
两边除以 2
2u=1
两边除以 222u​=21​
化简u=21​
u=21​
解为u=0,u=−21​,u=21​
u=sin(x)代回sin(x)=0,sin(x)=−21​,sin(x)=21​
sin(x)=0,sin(x)=−21​,sin(x)=21​
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
sin(x)=0的通解
sin(x) 周期表(周期为 2πn"):
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
解 x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
sin(x)=−21​:x=67π​+2πn,x=611π​+2πn
sin(x)=−21​
sin(x)=−21​的通解
sin(x) 周期表(周期为 2πn"):
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=67π​+2πn,x=611π​+2πn
x=67π​+2πn,x=611π​+2πn
sin(x)=21​:x=6π​+2πn,x=65π​+2πn
sin(x)=21​
sin(x)=21​的通解
sin(x) 周期表(周期为 2πn"):
x06π​4π​3π​2π​32π​43π​65π​​sin(x)021​22​​23​​123​​22​​21​​xπ67π​45π​34π​23π​35π​47π​611π​​sin(x)0−21​−22​​−23​​−1−23​​−22​​−21​​​
x=6π​+2πn,x=65π​+2πn
x=6π​+2πn,x=65π​+2πn
合并所有解x=2πn,x=π+2πn,x=67π​+2πn,x=611π​+2πn,x=6π​+2πn,x=65π​+2πn

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cot(θ)=-2/3cot(θ)=−32​cos(θ)= 1/7cos(θ)=71​sqrt(3)sin(2x)+3cos(2x)=03​sin(2x)+3cos(2x)=03cos(θ)=2sin^2(θ)3cos(θ)=2sin2(θ)sin(x)-sin(x)cos(x)=0sin(x)−sin(x)cos(x)=0
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