해법
cot(x)=sin2(x)
해법
x=0.97202…+2πn,x=π+0.97202…+2πn
+1
도
x=55.69319…∘+360∘n,x=235.69319…∘+360∘n솔루션 단계
cot(x)=sin2(x)
빼다 sin2(x) 양쪽에서cot(x)−sin2(x)=0
죄로 표현하라, 왜냐하면sin(x)cos(x)−sin2(x)=0
sin(x)cos(x)−sin2(x)단순화하세요:sin(x)cos(x)−sin3(x)
sin(x)cos(x)−sin2(x)
요소를 분수로 변환: sin2(x)=sin(x)sin2(x)sin(x)=sin(x)cos(x)−sin(x)sin2(x)sin(x)
분모가 같기 때문에, 분수를 합친다: ca±cb=ca±b=sin(x)cos(x)−sin2(x)sin(x)
cos(x)−sin2(x)sin(x)=cos(x)−sin3(x)
cos(x)−sin2(x)sin(x)
sin2(x)sin(x)=sin3(x)
sin2(x)sin(x)
지수 규칙 적용: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=sin2+1(x)
숫자 추가: 2+1=3=sin3(x)
=cos(x)−sin3(x)
=sin(x)cos(x)−sin3(x)
sin(x)cos(x)−sin3(x)=0
g(x)f(x)=0⇒f(x)=0cos(x)−sin3(x)=0
더하다 sin3(x) 양쪽으로cos(x)=sin3(x)
양쪽을 제곱cos2(x)=(sin3(x))2
빼다 (sin3(x))2 양쪽에서cos2(x)−sin6(x)=0
삼각성을 사용하여 다시 쓰기
cos2(x)−sin6(x)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=1−sin2(x)−sin6(x)
1−sin2(x)−sin6(x)=0
대체로 해결
1−sin2(x)−sin6(x)=0
하게: sin(x)=u1−u2−u6=0
1−u2−u6=0:u=0.68232…,u=−0.68232…
1−u2−u6=0
표준 양식으로 작성 anxn+…+a1x+a0=0−u6−u2+1=0
다음으로 방정식 다시 쓰기 v=u2 그리고 v3=u6−v3−v+1=0
−v3−v+1=0해결 :v≈0.68232…
−v3−v+1=0
다음을 위한 하나의 솔루션 찾기 −v3−v+1=0 뉴턴-랩슨을 이용하여:v≈0.68232…
−v3−v+1=0
뉴턴-랩슨 근사 정의
f(v)=−v3−v+1
f′(v)찾다 :−3v2−1
dvd(−v3−v+1)
합계/차이 규칙 적용: (f±g)′=f′±g′=−dvd(v3)−dvdv+dvd(1)
dvd(v3)=3v2
dvd(v3)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=3v3−1
단순화=3v2
dvdv=1
dvdv
공통 도함수 적용: dvdv=1=1
dvd(1)=0
dvd(1)
상수의 도함수: dxd(a)=0=0
=−3v2−1+0
단순화=−3v2−1
렛 v0=1계산하다 vn+1 까지 Δvn+1<0.000001
v1=0.75:Δv1=0.25
f(v0)=−13−1+1=−1f′(v0)=−3⋅12−1=−4v1=0.75
Δv1=∣0.75−1∣=0.25Δv1=0.25
v2=0.68604…:Δv2=0.06395…
f(v1)=−0.753−0.75+1=−0.171875f′(v1)=−3⋅0.752−1=−2.6875v2=0.68604…
Δv2=∣0.68604…−0.75∣=0.06395…Δv2=0.06395…
v3=0.68233…:Δv3=0.00370…
f(v2)=−0.68604…3−0.68604…+1=−0.00894…f′(v2)=−3⋅0.68604…2−1=−2.41197…v3=0.68233…
Δv3=∣0.68233…−0.68604…∣=0.00370…Δv3=0.00370…
v4=0.68232…:Δv4=0.00001…
f(v3)=−0.68233…3−0.68233…+1=−0.00002…f′(v3)=−3⋅0.68233…2−1=−2.39676…v4=0.68232…
Δv4=∣0.68232…−0.68233…∣=0.00001…Δv4=0.00001…
v5=0.68232…:Δv5=1.18493E−10
f(v4)=−0.68232…3−0.68232…+1=−2.83995E−10f′(v4)=−3⋅0.68232…2−1=−2.39671…v5=0.68232…
Δv5=∣0.68232…−0.68232…∣=1.18493E−10Δv5=1.18493E−10
v≈0.68232…
긴 나눗셈 적용:v−0.68232…−v3−v+1=−v2−0.68232…v−1.46557…
−v2−0.68232…v−1.46557…≈0
다음을 위한 하나의 솔루션 찾기 −v2−0.68232…v−1.46557…=0 뉴턴-랩슨을 이용하여:솔루션 없음 v∈R
−v2−0.68232…v−1.46557…=0
뉴턴-랩슨 근사 정의
f(v)=−v2−0.68232…v−1.46557…
f′(v)찾다 :−2v−0.68232…
dvd(−v2−0.68232…v−1.46557…)
합계/차이 규칙 적용: (f±g)′=f′±g′=−dvd(v2)−dvd(0.68232…v)−dvd(1.46557…)
dvd(v2)=2v
dvd(v2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=2v2−1
단순화=2v
dvd(0.68232…v)=0.68232…
dvd(0.68232…v)
정수를 빼라: (a⋅f)′=a⋅f′=0.68232…dvdv
공통 도함수 적용: dvdv=1=0.68232…⋅1
단순화=0.68232…
dvd(1.46557…)=0
dvd(1.46557…)
상수의 도함수: dxd(a)=0=0
=−2v−0.68232…−0
단순화=−2v−0.68232…
렛 v0=−2계산하다 vn+1 까지 Δvn+1<0.000001
v1=−0.76391…:Δv1=1.23608…
f(v0)=−(−2)2−0.68232…(−2)−1.46557…=−4.10091…f′(v0)=−2(−2)−0.68232…=3.31767…v1=−0.76391…
Δv1=∣−0.76391…−(−2)∣=1.23608…Δv1=1.23608…
v2=1.04316…:Δv2=1.80707…
f(v1)=−(−0.76391…)2−0.68232…(−0.76391…)−1.46557…=−1.52789…f′(v1)=−2(−0.76391…)−0.68232…=0.84550…v2=1.04316…
Δv2=∣1.04316…−(−0.76391…)∣=1.80707…Δv2=1.80707…
v3=−0.13630…:Δv3=1.17946…
f(v2)=−1.04316…2−0.68232…⋅1.04316…−1.46557…=−3.26553…f′(v2)=−2⋅1.04316…−0.68232…=−2.76865…v3=−0.13630…
Δv3=∣−0.13630…−1.04316…∣=1.17946…Δv3=1.17946…
v4=−3.53171…:Δv4=3.39540…
f(v3)=−(−0.13630…)2−0.68232…(−0.13630…)−1.46557…=−1.39114…f′(v3)=−2(−0.13630…)−0.68232…=−0.40971…v4=−3.53171…
Δv4=∣−3.53171…−(−0.13630…)∣=3.39540…Δv4=3.39540…
v5=−1.72500…:Δv5=1.80670…
f(v4)=−(−3.53171…)2−0.68232…(−3.53171…)−1.46557…=−11.52876…f′(v4)=−2(−3.53171…)−0.68232…=6.38109…v5=−1.72500…
Δv5=∣−1.72500…−(−3.53171…)∣=1.80670…Δv5=1.80670…
v6=−0.54560…:Δv6=1.17939…
f(v5)=−(−1.72500…)2−0.68232…(−1.72500…)−1.46557…=−3.26419…f′(v5)=−2(−1.72500…)−0.68232…=2.76767…v6=−0.54560…
Δv6=∣−0.54560…−(−1.72500…)∣=1.17939…Δv6=1.17939…
v7=2.85625…:Δv7=3.40185…
f(v6)=−(−0.54560…)2−0.68232…(−0.54560…)−1.46557…=−1.39097…f′(v6)=−2(−0.54560…)−0.68232…=0.40888…v7=2.85625…
Δv7=∣2.85625…−(−0.54560…)∣=3.40185…Δv7=3.40185…
v8=1.04656…:Δv8=1.80968…
f(v7)=−2.85625…2−0.68232…⋅2.85625…−1.46557…=−11.57264…f′(v7)=−2⋅2.85625…−0.68232…=−6.39483…v8=1.04656…
Δv8=∣1.04656…−2.85625…∣=1.80968…Δv8=1.80968…
v9=−0.13340…:Δv9=1.17997…
f(v8)=−1.04656…2−0.68232…⋅1.04656…−1.46557…=−3.27496…f′(v8)=−2⋅1.04656…−0.68232…=−2.77545…v9=−0.13340…
Δv9=∣−0.13340…−1.04656…∣=1.17997…Δv9=1.17997…
v10=−3.48434…:Δv10=3.35093…
f(v9)=−(−0.13340…)2−0.68232…(−0.13340…)−1.46557…=−1.39234…f′(v9)=−2(−0.13340…)−0.68232…=−0.41550…v10=−3.48434…
Δv10=∣−3.48434…−(−0.13340…)∣=3.35093…Δv10=3.35093…
해결 방법을 찾을 수 없습니다
해결책은v≈0.68232…
v≈0.68232…
다시 대체 v=u2,을 해결하다 u
u2=0.68232…해결 :u=0.68232…,u=−0.68232…
u2=0.68232…
위해서 x2=f(a) 해결책은 x=f(a),−f(a)
u=0.68232…,u=−0.68232…
해결책은
u=0.68232…,u=−0.68232…
뒤로 대체 u=sin(x)sin(x)=0.68232…,sin(x)=−0.68232…
sin(x)=0.68232…,sin(x)=−0.68232…
sin(x)=0.68232…:x=arcsin(0.68232…)+2πn,x=π−arcsin(0.68232…)+2πn
sin(x)=0.68232…
트리거 역속성 적용
sin(x)=0.68232…
일반 솔루션 sin(x)=0.68232…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.68232…)+2πn,x=π−arcsin(0.68232…)+2πn
x=arcsin(0.68232…)+2πn,x=π−arcsin(0.68232…)+2πn
sin(x)=−0.68232…:x=arcsin(−0.68232…)+2πn,x=π+arcsin(0.68232…)+2πn
sin(x)=−0.68232…
트리거 역속성 적용
sin(x)=−0.68232…
일반 솔루션 sin(x)=−0.68232…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.68232…)+2πn,x=π+arcsin(0.68232…)+2πn
x=arcsin(−0.68232…)+2πn,x=π+arcsin(0.68232…)+2πn
모든 솔루션 결합x=arcsin(0.68232…)+2πn,x=π−arcsin(0.68232…)+2πn,x=arcsin(−0.68232…)+2πn,x=π+arcsin(0.68232…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 cot(x)=sin2(x)
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arcsin(0.68232…)+2πn:참
arcsin(0.68232…)+2πn
n=1끼우다 arcsin(0.68232…)+2π1
cot(x)=sin2(x) 위한 {\ quad}끼우다{\ quad} x=arcsin(0.68232…)+2π1cot(arcsin(0.68232…)+2π1)=sin2(arcsin(0.68232…)+2π1)
다듬다0.68232…=0.68232…
⇒참
솔루션 확인 π−arcsin(0.68232…)+2πn:거짓
π−arcsin(0.68232…)+2πn
n=1끼우다 π−arcsin(0.68232…)+2π1
cot(x)=sin2(x) 위한 {\ quad}끼우다{\ quad} x=π−arcsin(0.68232…)+2π1cot(π−arcsin(0.68232…)+2π1)=sin2(π−arcsin(0.68232…)+2π1)
다듬다−0.68232…=0.68232…
⇒거짓
솔루션 확인 arcsin(−0.68232…)+2πn:거짓
arcsin(−0.68232…)+2πn
n=1끼우다 arcsin(−0.68232…)+2π1
cot(x)=sin2(x) 위한 {\ quad}끼우다{\ quad} x=arcsin(−0.68232…)+2π1cot(arcsin(−0.68232…)+2π1)=sin2(arcsin(−0.68232…)+2π1)
다듬다−0.68232…=0.68232…
⇒거짓
솔루션 확인 π+arcsin(0.68232…)+2πn:참
π+arcsin(0.68232…)+2πn
n=1끼우다 π+arcsin(0.68232…)+2π1
cot(x)=sin2(x) 위한 {\ quad}끼우다{\ quad} x=π+arcsin(0.68232…)+2π1cot(π+arcsin(0.68232…)+2π1)=sin2(π+arcsin(0.68232…)+2π1)
다듬다0.68232…=0.68232…
⇒참
x=arcsin(0.68232…)+2πn,x=π+arcsin(0.68232…)+2πn
해를 10진수 형식으로 표시x=0.97202…+2πn,x=π+0.97202…+2πn