解答
cos4(a)+cos2(a)=1
解答
a=0.66623…+2πn,a=2π−0.66623…+2πn,a=2.47535…+2πn,a=−2.47535…+2πn
+1
度数
a=38.17270…∘+360∘n,a=321.82729…∘+360∘n,a=141.82729…∘+360∘n,a=−141.82729…∘+360∘n求解步骤
cos4(a)+cos2(a)=1
用替代法求解
cos4(a)+cos2(a)=1
令:cos(a)=uu4+u2=1
u4+u2=1:u=2−1+5,u=−2−1+5,u=2−1−5,u=−2−1−5
u4+u2=1
将 1para o lado esquerdo
u4+u2=1
两边减去 1u4+u2−1=1−1
化简u4+u2−1=0
u4+u2−1=0
用v=u2 和 v2=u4改写方程式v2+v−1=0
解 v2+v−1=0:v=2−1+5,v=2−1−5
v2+v−1=0
使用求根公式求解
v2+v−1=0
二次方程求根公式:
若 a=1,b=1,c=−1v1,2=2⋅1−1±12−4⋅1⋅(−1)
v1,2=2⋅1−1±12−4⋅1⋅(−1)
12−4⋅1⋅(−1)=5
12−4⋅1⋅(−1)
使用法则 1a=112=1=1−4⋅1⋅(−1)
使用法则 −(−a)=a=1+4⋅1⋅1
数字相乘:4⋅1⋅1=4=1+4
数字相加:1+4=5=5
v1,2=2⋅1−1±5
将解分隔开v1=2⋅1−1+5,v2=2⋅1−1−5
v=2⋅1−1+5:2−1+5
2⋅1−1+5
数字相乘:2⋅1=2=2−1+5
v=2⋅1−1−5:2−1−5
2⋅1−1−5
数字相乘:2⋅1=2=2−1−5
二次方程组的解是:v=2−1+5,v=2−1−5
v=2−1+5,v=2−1−5
代回 v=u2，求解 u
解 u2=2−1+5:u=2−1+5,u=−2−1+5
u2=2−1+5
对于 x2=f(a) 解为 x=f(a),−f(a)
u=2−1+5,u=−2−1+5
解 u2=2−1−5:u=2−1−5,u=−2−1−5
u2=2−1−5
对于 x2=f(a) 解为 x=f(a),−f(a)
u=2−1−5,u=−2−1−5
解为
u=2−1+5,u=−2−1+5,u=2−1−5,u=−2−1−5
u=cos(a)代回cos(a)=2−1+5,cos(a)=−2−1+5,cos(a)=2−1−5,cos(a)=−2−1−5
cos(a)=2−1+5,cos(a)=−2−1+5,cos(a)=2−1−5,cos(a)=−2−1−5
cos(a)=2−1+5:a=arccos2−1+5+2πn,a=2π−arccos2−1+5+2πn
cos(a)=2−1+5
使用反三角函数性质
cos(a)=2−1+5
cos(a)=2−1+5的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πna=arccos2−1+5+2πn,a=2π−arccos2−1+5+2πn
a=arccos2−1+5+2πn,a=2π−arccos2−1+5+2πn
cos(a)=−2−1+5:a=arccos−2−1+5+2πn,a=−arccos−2−1+5+2πn
cos(a)=−2−1+5
使用反三角函数性质
cos(a)=−2−1+5
cos(a)=−2−1+5的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πna=arccos−2−1+5+2πn,a=−arccos−2−1+5+2πn
a=arccos−2−1+5+2πn,a=−arccos−2−1+5+2πn
cos(a)=2−1−5:a=arccos2−1−5+2πn,a=−arccos2−1−5+2πn
cos(a)=2−1−5
使用反三角函数性质
cos(a)=2−1−5
cos(a)=2−1−5的通解cos(x)=a⇒x=arccos(a)+2πn,x=−arccos(a)+2πna=arccos2−1−5+2πn,a=−arccos2−1−5+2πn
a=arccos2−1−5+2πn,a=−arccos2−1−5+2πn
cos(a)=−2−1−5:a=arccos−2−1−5+2πn,a=−arccos−2−1−5+2πn
cos(a)=−2−1−5
使用反三角函数性质
cos(a)=−2−1−5
cos(a)=−2−1−5的通解cos(x)=a⇒x=arccos(a)+2πn,x=−arccos(a)+2πna=arccos−2−1−5+2πn,a=−arccos−2−1−5+2πn
a=arccos−2−1−5+2πn,a=−arccos−2−1−5+2πn
合并所有解a=arccos2−1+5+2πn,a=2π−arccos2−1+5+2πn,a=arccos−2−1+5+2πn,a=−arccos−2−1+5+2πn,a=arccos2−1−5+2πn,a=−arccos2−1−5+2πn,a=arccos−2−1−5+2πn,a=−arccos−2−1−5+2πn
因为方程对以下值无定义:arccos2−1−5+2πn,−arccos2−1−5+2πn,arccos−2−1−5+2πn,−arccos−2−1−5+2πna=arccos2−1+5+2πn,a=2π−arccos2−1+5+2πn,a=arccos−2−1+5+2πn,a=−arccos−2−1+5+2πn
以小数形式表示解a=0.66623…+2πn,a=2π−0.66623…+2πn,a=2.47535…+2πn,a=−2.47535…+2πn