解答
求解 n,sin(x)+sin(132n−x)=1
解答
n=132arcsin(1−sin(x))+134πk+132x,n=132π+132arcsin(−1+sin(x))+134πk+132x
求解步骤
sin(x)+sin(13⋅2n−x)=1
将 sin(x)到右边
sin(x)+sin(132n−x)=1
两边减去 sin(x)sin(x)+sin(132n−x)−sin(x)=1−sin(x)
化简sin(132n−x)=1−sin(x)
sin(132n−x)=1−sin(x)
使用反三角函数性质
sin(13⋅2n−x)=1−sin(x)
sin(132n−x)=1−sin(x)的通解sin(x)=a⇒x=arcsin(a)+2πk,x=π+arcsin(a)+2πk13⋅2n−x=arcsin(1−sin(x))+2πk,13⋅2n−x=π+arcsin(−1+sin(x))+2πk
13⋅2n−x=arcsin(1−sin(x))+2πk,13⋅2n−x=π+arcsin(−1+sin(x))+2πk
解 13⋅2n−x=arcsin(1−sin(x))+2πk:n=132arcsin(1−sin(x))+134πk+132x
13⋅2n−x=arcsin(1−sin(x))+2πk
将 x到右边
13⋅2n−x=arcsin(1−sin(x))+2πk
两边加上 x13⋅2n−x+x=arcsin(1−sin(x))+2πk+x
化简13⋅2n=arcsin(1−sin(x))+2πk+x
13⋅2n=arcsin(1−sin(x))+2πk+x
整理 13⋅2n:213n
13⋅2n
分式相乘: a⋅cb=ca⋅b=2n⋅13
213n=arcsin(1−sin(x))+2πk+x
在两边乘以 2
213n=arcsin(1−sin(x))+2πk+x
在两边乘以 2213n⋅2=arcsin(1−sin(x))⋅2+2πk⋅2+x⋅2
化简
213n⋅2=arcsin(1−sin(x))⋅2+2πk⋅2+x⋅2
化简 213n⋅2:13n
213n⋅2
分式相乘: a⋅cb=ca⋅b=213n⋅2
约分:2=13n
化简 arcsin(1−sin(x))⋅2:2arcsin(1−sin(x))
arcsin(1−sin(x))⋅2
使用交换律:arcsin(1−sin(x))⋅2=2arcsin(1−sin(x))2arcsin(1−sin(x))
化简 2πk⋅2:4πk
2πk⋅2
数字相乘:2⋅2=4=4πk
化简 x⋅2:2x
x⋅2
使用交换律:x⋅2=2x2x
13n=2arcsin(1−sin(x))+4πk+2x
13n=2arcsin(1−sin(x))+4πk+2x
13n=2arcsin(1−sin(x))+4πk+2x
两边除以 13
13n=2arcsin(1−sin(x))+4πk+2x
两边除以 131313n=132arcsin(1−sin(x))+134πk+132x
化简n=132arcsin(1−sin(x))+134πk+132x
n=132arcsin(1−sin(x))+134πk+132x
解 13⋅2n−x=π+arcsin(−1+sin(x))+2πk:n=132π+132arcsin(−1+sin(x))+134πk+132x
13⋅2n−x=π+arcsin(−1+sin(x))+2πk
将 x到右边
13⋅2n−x=π+arcsin(−1+sin(x))+2πk
两边加上 x13⋅2n−x+x=π+arcsin(−1+sin(x))+2πk+x
化简13⋅2n=π+arcsin(−1+sin(x))+2πk+x
13⋅2n=π+arcsin(−1+sin(x))+2πk+x
整理 13⋅2n:213n
13⋅2n
分式相乘: a⋅cb=ca⋅b=2n⋅13
213n=π+arcsin(−1+sin(x))+2πk+x
在两边乘以 2
213n=π+arcsin(−1+sin(x))+2πk+x
在两边乘以 2213n⋅2=π2+arcsin(−1+sin(x))⋅2+2πk⋅2+x⋅2
化简
213n⋅2=π2+arcsin(−1+sin(x))⋅2+2πk⋅2+x⋅2
化简 213n⋅2:13n
213n⋅2
分式相乘: a⋅cb=ca⋅b=213n⋅2
约分:2=13n
化简 π2:2π
π2
使用交换律:π2=2π2π
化简 arcsin(−1+sin(x))⋅2:2arcsin(−1+sin(x))
arcsin(−1+sin(x))⋅2
使用交换律:arcsin(−1+sin(x))⋅2=2arcsin(−1+sin(x))2arcsin(−1+sin(x))
化简 2πk⋅2:4πk
2πk⋅2
数字相乘:2⋅2=4=4πk
化简 x⋅2:2x
x⋅2
使用交换律:x⋅2=2x2x
13n=2π+2arcsin(−1+sin(x))+4πk+2x
13n=2π+2arcsin(−1+sin(x))+4πk+2x
13n=2π+2arcsin(−1+sin(x))+4πk+2x
两边除以 13
13n=2π+2arcsin(−1+sin(x))+4πk+2x
两边除以 131313n=132π+132arcsin(−1+sin(x))+134πk+132x
化简n=132π+132arcsin(−1+sin(x))+134πk+132x
n=132π+132arcsin(−1+sin(x))+134πk+132x
n=132arcsin(1−sin(x))+134πk+132x,n=132π+132arcsin(−1+sin(x))+134πk+132x