解答
2cos(x)=2cos(3x)
解答
x=2π+2πn,x=23π+2πn,x=π+2πn,x=2πn
+1
度数
x=90∘+360∘n,x=270∘+360∘n,x=180∘+360∘n,x=0∘+360∘n求解步骤
2cos(x)=2cos(3x)
两边减去 2cos(3x)2cos(x)−2cos(3x)=0
使用三角恒等式改写
−2cos(3x)+2cos(x)
cos(3x)=4cos3(x)−3cos(x)
cos(3x)
使用三角恒等式改写
cos(3x)
改写为=cos(2x+x)
使用角和恒等式: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(2x)cos(x)−sin(2x)sin(x)
使用倍角公式: sin(2x)=2sin(x)cos(x)=cos(2x)cos(x)−2sin(x)cos(x)sin(x)
化简 cos(2x)cos(x)−2sin(x)cos(x)sin(x):cos(x)cos(2x)−2sin2(x)cos(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)
2sin(x)cos(x)sin(x)=2sin2(x)cos(x)
2sin(x)cos(x)sin(x)
使用指数法则: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=2cos(x)sin1+1(x)
数字相加:1+1=2=2cos(x)sin2(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
使用倍角公式: cos(2x)=2cos2(x)−1=(2cos2(x)−1)cos(x)−2sin2(x)cos(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
乘开 (2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x):4cos3(x)−3cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
=cos(x)(2cos2(x)−1)−2cos(x)(1−cos2(x))
乘开 cos(x)(2cos2(x)−1):2cos3(x)−cos(x)
cos(x)(2cos2(x)−1)
使用分配律: a(b−c)=ab−aca=cos(x),b=2cos2(x),c=1=cos(x)2cos2(x)−cos(x)1
=2cos2(x)cos(x)−1cos(x)
化简 2cos2(x)cos(x)−1⋅cos(x):2cos3(x)−cos(x)
2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
数字相加:2+1=3=2cos3(x)
1⋅cos(x)=cos(x)
1cos(x)
乘以:1⋅cos(x)=cos(x)=cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)−2(1−cos2(x))cos(x)
乘开 −2cos(x)(1−cos2(x)):−2cos(x)+2cos3(x)
−2cos(x)(1−cos2(x))
使用分配律: a(b−c)=ab−aca=−2cos(x),b=1,c=cos2(x)=−2cos(x)1−(−2cos(x))cos2(x)
使用加减运算法则−(−a)=a=−2⋅1cos(x)+2cos2(x)cos(x)
化简 −2⋅1⋅cos(x)+2cos2(x)cos(x):−2cos(x)+2cos3(x)
−2⋅1cos(x)+2cos2(x)cos(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1cos(x)
数字相乘:2⋅1=2=2cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
数字相加:2+1=3=2cos3(x)
=−2cos(x)+2cos3(x)
=−2cos(x)+2cos3(x)
=2cos3(x)−cos(x)−2cos(x)+2cos3(x)
化简 2cos3(x)−cos(x)−2cos(x)+2cos3(x):4cos3(x)−3cos(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)
对同类项分组=2cos3(x)+2cos3(x)−cos(x)−2cos(x)
同类项相加:2cos3(x)+2cos3(x)=4cos3(x)=4cos3(x)−cos(x)−2cos(x)
同类项相加:−cos(x)−2cos(x)=−3cos(x)=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=−2(4cos3(x)−3cos(x))+2cos(x)
化简 −2(4cos3(x)−3cos(x))+2cos(x):−8cos3(x)+8cos(x)
−2(4cos3(x)−3cos(x))+2cos(x)
乘开 −2(4cos3(x)−3cos(x)):−8cos3(x)+6cos(x)
−2(4cos3(x)−3cos(x))
使用分配律: a(b−c)=ab−aca=−2,b=4cos3(x),c=3cos(x)=−2⋅4cos3(x)−(−2)⋅3cos(x)
使用加减运算法则−(−a)=a=−2⋅4cos3(x)+2⋅3cos(x)
化简 −2⋅4cos3(x)+2⋅3cos(x):−8cos3(x)+6cos(x)
−2⋅4cos3(x)+2⋅3cos(x)
数字相乘:2⋅4=8=−8cos3(x)+2⋅3cos(x)
数字相乘:2⋅3=6=−8cos3(x)+6cos(x)
=−8cos3(x)+6cos(x)
=−8cos3(x)+6cos(x)+2cos(x)
同类项相加:6cos(x)+2cos(x)=8cos(x)=−8cos3(x)+8cos(x)
=−8cos3(x)+8cos(x)
8cos(x)−8cos3(x)=0
用替代法求解
8cos(x)−8cos3(x)=0
令:cos(x)=u8u−8u3=0
8u−8u3=0:u=0,u=−1,u=1
8u−8u3=0
因式分解 8u−8u3:−8u(u+1)(u−1)
8u−8u3
因式分解出通项 −8u:−8u(u2−1)
−8u3+8u
使用指数法则: ab+c=abacu3=u2u=−8u2u+8u
因式分解出通项 −8u=−8u(u2−1)
=−8u(u2−1)
分解 u2−1:(u+1)(u−1)
u2−1
将 1 改写为 12=u2−12
使用平方差公式: x2−y2=(x+y)(x−y)u2−12=(u+1)(u−1)=(u+1)(u−1)
=−8u(u+1)(u−1)
−8u(u+1)(u−1)=0
使用零因数法则: If ab=0then a=0or b=0u=0oru+1=0oru−1=0
解 u+1=0:u=−1
u+1=0
将 1到右边
u+1=0
两边减去 1u+1−1=0−1
化简u=−1
u=−1
解 u−1=0:u=1
u−1=0
将 1到右边
u−1=0
两边加上 1u−1+1=0+1
化简u=1
u=1
解为u=0,u=−1,u=1
u=cos(x)代回cos(x)=0,cos(x)=−1,cos(x)=1
cos(x)=0,cos(x)=−1,cos(x)=1
cos(x)=0:x=2π+2πn,x=23π+2πn
cos(x)=0
cos(x)=0的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=2π+2πn,x=23π+2πn
x=2π+2πn,x=23π+2πn
cos(x)=−1:x=π+2πn
cos(x)=−1
cos(x)=−1的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=π+2πn
x=π+2πn
cos(x)=1:x=2πn
cos(x)=1
cos(x)=1的通解
cos(x) 周期表(周期为 2πn):
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=0+2πn
x=0+2πn
解 x=0+2πn:x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn
合并所有解x=2π+2πn,x=23π+2πn,x=π+2πn,x=2πn