解答
14sin(x+2π)+21tan(π−x)=0
解答
x=6π+2πn,x=65π+2πn
+1
度数
x=30∘+360∘n,x=150∘+360∘n求解步骤
14sin(x+2π)+21tan(π−x)=0
使用三角恒等式改写
14sin(x+2π)+21tan(π−x)=0
使用三角恒等式改写
sin(x+2π)
使用角和恒等式: sin(s+t)=sin(s)cos(t)+cos(s)sin(t)=sin(x)cos(2π)+cos(x)sin(2π)
化简 sin(x)cos(2π)+cos(x)sin(2π):cos(x)
sin(x)cos(2π)+cos(x)sin(2π)
sin(x)cos(2π)=0
sin(x)cos(2π)
化简 cos(2π):0
cos(2π)
使用以下普通恒等式:cos(2π)=0
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=0=0⋅sin(x)
使用法则 0⋅a=0=0
cos(x)sin(2π)=cos(x)
cos(x)sin(2π)
化简 sin(2π):1
sin(2π)
使用以下普通恒等式:sin(2π)=1
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=1=1⋅cos(x)
乘以:cos(x)⋅1=cos(x)=cos(x)
=0+cos(x)
0+cos(x)=cos(x)=cos(x)
=cos(x)
使用基本三角恒等式: tan(x)=cos(x)sin(x)=cos(π−x)sin(π−x)
使用角差恒等式: sin(s−t)=sin(s)cos(t)−cos(s)sin(t)=cos(π−x)sin(π)cos(x)−cos(π)sin(x)
使用角差恒等式: cos(s−t)=cos(s)cos(t)+sin(s)sin(t)=cos(π)cos(x)+sin(π)sin(x)sin(π)cos(x)−cos(π)sin(x)
化简 cos(π)cos(x)+sin(π)sin(x)sin(π)cos(x)−cos(π)sin(x):−cos(x)sin(x)
cos(π)cos(x)+sin(π)sin(x)sin(π)cos(x)−cos(π)sin(x)
sin(π)cos(x)−cos(π)sin(x)=sin(x)
sin(π)cos(x)−cos(π)sin(x)
sin(π)cos(x)=0
sin(π)cos(x)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅cos(x)
使用法则 0⋅a=0=0
cos(π)sin(x)=−sin(x)
cos(π)sin(x)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=−1⋅sin(x)
乘以:1⋅sin(x)=sin(x)=−sin(x)
=0−(−sin(x))
整理后得=sin(x)
=cos(π)cos(x)+sin(π)sin(x)sin(x)
cos(π)cos(x)+sin(π)sin(x)=−cos(x)
cos(π)cos(x)+sin(π)sin(x)
cos(π)cos(x)=−cos(x)
cos(π)cos(x)
化简 cos(π):−1
cos(π)
使用以下普通恒等式:cos(π)=(−1)
cos(x) 周期表（周期为 2πn）：
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
=−1=−1⋅cos(x)
乘以:1⋅cos(x)=cos(x)=−cos(x)
=−cos(x)+sin(π)sin(x)
sin(π)sin(x)=0
sin(π)sin(x)
化简 sin(π):0
sin(π)
使用以下普通恒等式:sin(π)=0
sin(x) 周期表（周期为 2πn"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
=0=0⋅sin(x)
使用法则 0⋅a=0=0
=−cos(x)+0
−cos(x)+0=−cos(x)=−cos(x)
=−cos(x)sin(x)
使用分式法则: −ba=−ba=−cos(x)sin(x)
=−cos(x)sin(x)
14cos(x)+21(−cos(x)sin(x))=0
化简 14cos(x)+21(−cos(x)sin(x)):14cos(x)−cos(x)21sin(x)
14cos(x)+21(−cos(x)sin(x))
去除括号: (−a)=−a=14cos(x)−21⋅cos(x)sin(x)
乘 21⋅cos(x)sin(x):cos(x)21sin(x)
21⋅cos(x)sin(x)
分式相乘: a⋅cb=ca⋅b=cos(x)sin(x)⋅21
=14cos(x)−cos(x)21sin(x)
14cos(x)−cos(x)21sin(x)=0
14cos(x)−cos(x)21sin(x)=0
化简 14cos(x)−cos(x)21sin(x):cos(x)14cos2(x)−21sin(x)
14cos(x)−cos(x)21sin(x)
将项转换为分式: 14cos(x)=cos(x)14cos(x)cos(x)=cos(x)14cos(x)cos(x)−cos(x)21sin(x)
因为分母相等，所以合并分式: ca±cb=ca±b=cos(x)14cos(x)cos(x)−21sin(x)
14cos(x)cos(x)−21sin(x)=14cos2(x)−21sin(x)
14cos(x)cos(x)−21sin(x)
14cos(x)cos(x)=14cos2(x)
14cos(x)cos(x)
使用指数法则: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=14cos1+1(x)
数字相加:1+1=2=14cos2(x)
=14cos2(x)−21sin(x)
=cos(x)14cos2(x)−21sin(x)
cos(x)14cos2(x)−21sin(x)=0
g(x)f(x)=0⇒f(x)=014cos2(x)−21sin(x)=0
使用三角恒等式改写
14cos2(x)−21sin(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=14(1−sin2(x))−21sin(x)
(1−sin2(x))⋅14−21sin(x)=0
用替代法求解
(1−sin2(x))⋅14−21sin(x)=0
令:sin(x)=u(1−u2)⋅14−21u=0
(1−u2)⋅14−21u=0:u=−2,u=21
(1−u2)⋅14−21u=0
展开 (1−u2)⋅14−21u:14−14u2−21u
(1−u2)⋅14−21u
=14(1−u2)−21u
乘开 14(1−u2):14−14u2
14(1−u2)
使用分配律: a(b−c)=ab−aca=14,b=1,c=u2=14⋅1−14u2
数字相乘:14⋅1=14=14−14u2
=14−14u2−21u
14−14u2−21u=0
改写成标准形式 ax2+bx+c=0−14u2−21u+14=0
使用求根公式求解
−14u2−21u+14=0
二次方程求根公式:
若 a=−14,b=−21,c=14u1,2=2(−14)−(−21)±(−21)2−4(−14)⋅14
u1,2=2(−14)−(−21)±(−21)2−4(−14)⋅14
(−21)2−4(−14)⋅14=35
(−21)2−4(−14)⋅14
使用法则 −(−a)=a=(−21)2+4⋅14⋅14
使用指数法则: (−a)n=an,若 n 是偶数(−21)2=212=212+4⋅14⋅14
数字相乘:4⋅14⋅14=784=212+784
212=441=441+784
数字相加:441+784=1225=1225
因式分解数字: 1225=352=352
使用根式运算法则: nan=a352=35=35
u1,2=2(−14)−(−21)±35
将解分隔开u1=2(−14)−(−21)+35,u2=2(−14)−(−21)−35
u=2(−14)−(−21)+35:−2
2(−14)−(−21)+35
去除括号: (−a)=−a,−(−a)=a=−2⋅1421+35
数字相加:21+35=56=−2⋅1456
数字相乘:2⋅14=28=−2856
使用分式法则: −ba=−ba=−2856
数字相除:2856=2=−2
u=2(−14)−(−21)−35:21
2(−14)−(−21)−35
去除括号: (−a)=−a,−(−a)=a=−2⋅1421−35
数字相减:21−35=−14=−2⋅14−14
数字相乘:2⋅14=28=−28−14
使用分式法则: −b−a=ba=2814
约分:14=21
二次方程组的解是:u=−2,u=21
u=sin(x)代回sin(x)=−2,sin(x)=21
sin(x)=−2,sin(x)=21
sin(x)=−2:无解
sin(x)=−2
−1≤sin(x)≤1无解
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"）：
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(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=6π+2πn,x=65π+2πn