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受欢迎的 三角函数 >

2cos^2(x)+tan^2(x)=2

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解答

2cos2(x)+tan2(x)=2

解答

x=2πn,x=π+2πn,x=43π​+2πn,x=45π​+2πn,x=4π​+2πn,x=47π​+2πn
+1
度数
x=0∘+360∘n,x=180∘+360∘n,x=135∘+360∘n,x=225∘+360∘n,x=45∘+360∘n,x=315∘+360∘n
求解步骤
2cos2(x)+tan2(x)=2
两边减去 22cos2(x)+tan2(x)−2=0
使用三角恒等式改写
−2+tan2(x)+2cos2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−2+tan2(x)+2(1−sin2(x))
化简 −2+tan2(x)+2(1−sin2(x)):tan2(x)−2sin2(x)
−2+tan2(x)+2(1−sin2(x))
乘开 2(1−sin2(x)):2−2sin2(x)
2(1−sin2(x))
使用分配律: a(b−c)=ab−aca=2,b=1,c=sin2(x)=2⋅1−2sin2(x)
数字相乘:2⋅1=2=2−2sin2(x)
=−2+tan2(x)+2−2sin2(x)
化简 −2+tan2(x)+2−2sin2(x):tan2(x)−2sin2(x)
−2+tan2(x)+2−2sin2(x)
对同类项分组=tan2(x)−2sin2(x)−2+2
−2+2=0=tan2(x)−2sin2(x)
=tan2(x)−2sin2(x)
=tan2(x)−2sin2(x)
tan2(x)−2sin2(x)=0
分解 tan2(x)−2sin2(x):(tan(x)+2​sin(x))(tan(x)−2​sin(x))
tan2(x)−2sin2(x)
将 tan2(x)−2sin2(x) 改写为 tan2(x)−(2​sin(x))2
tan2(x)−2sin2(x)
使用根式运算法则: a=(a​)22=(2​)2=tan2(x)−(2​)2sin2(x)
使用指数法则: ambm=(ab)m(2​)2sin2(x)=(2​sin(x))2=tan2(x)−(2​sin(x))2
=tan2(x)−(2​sin(x))2
使用平方差公式: x2−y2=(x+y)(x−y)tan2(x)−(2​sin(x))2=(tan(x)+2​sin(x))(tan(x)−2​sin(x))=(tan(x)+2​sin(x))(tan(x)−2​sin(x))
(tan(x)+sin(x)2​)(tan(x)−sin(x)2​)=0
分别求解每个部分tan(x)+sin(x)2​=0ortan(x)−sin(x)2​=0
tan(x)+sin(x)2​=0:x=2πn,x=π+2πn,x=43π​+2πn,x=45π​+2πn
tan(x)+sin(x)2​=0
用 sin, cos 表示
tan(x)+sin(x)2​
使用基本三角恒等式: tan(x)=cos(x)sin(x)​=cos(x)sin(x)​+sin(x)2​
化简 cos(x)sin(x)​+sin(x)2​:cos(x)sin(x)+2​sin(x)cos(x)​
cos(x)sin(x)​+sin(x)2​
将项转换为分式: 2​sin(x)=cos(x)sin(x)2​cos(x)​=cos(x)sin(x)​+cos(x)sin(x)2​cos(x)​
因为分母相等,所以合并分式: ca​±cb​=ca±b​=cos(x)sin(x)+sin(x)2​cos(x)​
=cos(x)sin(x)+2​sin(x)cos(x)​
cos(x)sin(x)+cos(x)sin(x)2​​=0
g(x)f(x)​=0⇒f(x)=0sin(x)+cos(x)sin(x)2​=0
分解 sin(x)+cos(x)sin(x)2​:sin(x)(1+2​cos(x))
sin(x)+cos(x)sin(x)2​
因式分解出通项 sin(x)=sin(x)(1+cos(x)2​)
sin(x)(1+2​cos(x))=0
分别求解每个部分sin(x)=0or1+2​cos(x)=0
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
1+2​cos(x)=0:x=43π​+2πn,x=45π​+2πn
1+2​cos(x)=0
将 1到右边
1+2​cos(x)=0
两边减去 11+2​cos(x)−1=0−1
化简2​cos(x)=−1
2​cos(x)=−1
两边除以 2​
2​cos(x)=−1
两边除以 2​2​2​cos(x)​=2​−1​
化简
2​2​cos(x)​=2​−1​
化简 2​2​cos(x)​:cos(x)
2​2​cos(x)​
约分:2​=cos(x)
化简 2​−1​:−22​​
2​−1​
使用分式法则: b−a​=−ba​=−2​1​
−2​1​有理化:−22​​
−2​1​
乘以共轭根式 2​2​​=−2​2​1⋅2​​
1⋅2​=2​
2​2​=2
2​2​
使用根式运算法则: a​a​=a2​2​=2=2
=−22​​
=−22​​
cos(x)=−22​​
cos(x)=−22​​
cos(x)=−22​​
cos(x)=−22​​的通解
cos(x) 周期表(周期为 2πn):
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=43π​+2πn,x=45π​+2πn
x=43π​+2πn,x=45π​+2πn
合并所有解x=2πn,x=π+2πn,x=43π​+2πn,x=45π​+2πn
tan(x)−sin(x)2​=0:x=2πn,x=π+2πn,x=4π​+2πn,x=47π​+2πn
tan(x)−sin(x)2​=0
用 sin, cos 表示
tan(x)−sin(x)2​
使用基本三角恒等式: tan(x)=cos(x)sin(x)​=cos(x)sin(x)​−sin(x)2​
化简 cos(x)sin(x)​−sin(x)2​:cos(x)sin(x)−2​sin(x)cos(x)​
cos(x)sin(x)​−sin(x)2​
将项转换为分式: 2​sin(x)=cos(x)sin(x)2​cos(x)​=cos(x)sin(x)​−cos(x)sin(x)2​cos(x)​
因为分母相等,所以合并分式: ca​±cb​=ca±b​=cos(x)sin(x)−sin(x)2​cos(x)​
=cos(x)sin(x)−2​sin(x)cos(x)​
cos(x)sin(x)−cos(x)sin(x)2​​=0
g(x)f(x)​=0⇒f(x)=0sin(x)−cos(x)sin(x)2​=0
分解 sin(x)−cos(x)sin(x)2​:sin(x)(1−2​cos(x))
sin(x)−cos(x)sin(x)2​
因式分解出通项 sin(x)=sin(x)(1−cos(x)2​)
sin(x)(1−2​cos(x))=0
分别求解每个部分sin(x)=0or1−2​cos(x)=0
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
1−2​cos(x)=0:x=4π​+2πn,x=47π​+2πn
1−2​cos(x)=0
将 1到右边
1−2​cos(x)=0
两边减去 11−2​cos(x)−1=0−1
化简−2​cos(x)=−1
−2​cos(x)=−1
两边除以 −2​
−2​cos(x)=−1
两边除以 −2​−2​−2​cos(x)​=−2​−1​
化简
−2​−2​cos(x)​=−2​−1​
化简 −2​−2​cos(x)​:cos(x)
−2​−2​cos(x)​
使用分式法则: −b−a​=ba​=2​2​cos(x)​
约分:2​=cos(x)
化简 −2​−1​:22​​
−2​−1​
使用分式法则: −b−a​=ba​=2​1​
2​1​有理化:22​​
2​1​
乘以共轭根式 2​2​​=2​2​1⋅2​​
1⋅2​=2​
2​2​=2
2​2​
使用根式运算法则: a​a​=a2​2​=2=2
=22​​
=22​​
cos(x)=22​​
cos(x)=22​​
cos(x)=22​​
cos(x)=22​​的通解
cos(x) 周期表(周期为 2πn):
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=4π​+2πn,x=47π​+2πn
x=4π​+2πn,x=47π​+2πn
合并所有解x=2πn,x=π+2πn,x=4π​+2πn,x=47π​+2πn
合并所有解x=2πn,x=π+2πn,x=43π​+2πn,x=45π​+2πn,x=4π​+2πn,x=47π​+2πn

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