해법
sin3(x)−2sin(x)=4cos2(x)−3
해법
x=−0.32154…+2πn,x=π+0.32154…+2πn,x=0.80194…+2πn,x=π−0.80194…+2πn
+1
도
x=−18.42305…∘+360∘n,x=198.42305…∘+360∘n,x=45.94805…∘+360∘n,x=134.05194…∘+360∘n솔루션 단계
sin3(x)−2sin(x)=4cos2(x)−3
빼다 4cos2(x)−3 양쪽에서sin3(x)−2sin(x)−4cos2(x)+3=0
삼각성을 사용하여 다시 쓰기
3+sin3(x)−2sin(x)−4cos2(x)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=3+sin3(x)−2sin(x)−4(1−sin2(x))
3+sin3(x)−2sin(x)−4(1−sin2(x))간소화하다 :sin3(x)+4sin2(x)−2sin(x)−1
3+sin3(x)−2sin(x)−4(1−sin2(x))
−4(1−sin2(x))확대한다:−4+4sin2(x)
−4(1−sin2(x))
분배 법칙 적용: a(b−c)=ab−aca=−4,b=1,c=sin2(x)=−4⋅1−(−4)sin2(x)
마이너스 플러스 규칙 적용−(−a)=a=−4⋅1+4sin2(x)
숫자를 곱하시오: 4⋅1=4=−4+4sin2(x)
=3+sin3(x)−2sin(x)−4+4sin2(x)
3+sin3(x)−2sin(x)−4+4sin2(x)단순화하세요:sin3(x)+4sin2(x)−2sin(x)−1
3+sin3(x)−2sin(x)−4+4sin2(x)
집단적 용어=sin3(x)−2sin(x)+4sin2(x)+3−4
숫자 더하기/ 빼기: 3−4=−1=sin3(x)+4sin2(x)−2sin(x)−1
=sin3(x)+4sin2(x)−2sin(x)−1
=sin3(x)+4sin2(x)−2sin(x)−1
−1+sin3(x)−2sin(x)+4sin2(x)=0
대체로 해결
−1+sin3(x)−2sin(x)+4sin2(x)=0
하게: sin(x)=u−1+u3−2u+4u2=0
−1+u3−2u+4u2=0:u≈−0.31603…,u≈0.71870…,u≈−4.40267…
−1+u3−2u+4u2=0
표준 양식으로 작성 anxn+…+a1x+a0=0u3+4u2−2u−1=0
다음을 위한 하나의 솔루션 찾기 u3+4u2−2u−1=0 뉴턴-랩슨을 이용하여:u≈−0.31603…
u3+4u2−2u−1=0
뉴턴-랩슨 근사 정의
f(u)=u3+4u2−2u−1
f′(u)찾다 :3u2+8u−2
dud(u3+4u2−2u−1)
합계/차이 규칙 적용: (f±g)′=f′±g′=dud(u3)+dud(4u2)−dud(2u)−dud(1)
dud(u3)=3u2
dud(u3)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=3u3−1
단순화=3u2
dud(4u2)=8u
dud(4u2)
정수를 빼라: (a⋅f)′=a⋅f′=4dud(u2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=4⋅2u2−1
단순화=8u
dud(2u)=2
dud(2u)
정수를 빼라: (a⋅f)′=a⋅f′=2dudu
공통 도함수 적용: dudu=1=2⋅1
단순화=2
dud(1)=0
dud(1)
상수의 도함수: dxd(a)=0=0
=3u2+8u−2−0
단순화=3u2+8u−2
렛 u0=0계산하다 un+1 까지 Δun+1<0.000001
u1=−0.5:Δu1=0.5
f(u0)=03+4⋅02−2⋅0−1=−1f′(u0)=3⋅02+8⋅0−2=−2u1=−0.5
Δu1=∣−0.5−0∣=0.5Δu1=0.5
u2=−0.33333…:Δu2=0.16666…
f(u1)=(−0.5)3+4(−0.5)2−2(−0.5)−1=0.875f′(u1)=3(−0.5)2+8(−0.5)−2=−5.25u2=−0.33333…
Δu2=∣−0.33333…−(−0.5)∣=0.16666…Δu2=0.16666…
u3=−0.31623…:Δu3=0.01709…
f(u2)=(−0.33333…)3+4(−0.33333…)2−2(−0.33333…)−1=0.07407…f′(u2)=3(−0.33333…)2+8(−0.33333…)−2=−4.33333…u3=−0.31623…
Δu3=∣−0.31623…−(−0.33333…)∣=0.01709…Δu3=0.01709…
u4=−0.31603…:Δu4=0.00020…
f(u3)=(−0.31623…)3+4(−0.31623…)2−2(−0.31623…)−1=0.00088…f′(u3)=3(−0.31623…)2+8(−0.31623…)−2=−4.22989…u4=−0.31603…
Δu4=∣−0.31603…−(−0.31623…)∣=0.00020…Δu4=0.00020…
u5=−0.31603…:Δu5=3.13479E−8
f(u4)=(−0.31603…)3+4(−0.31603…)2−2(−0.31603…)−1=1.32558E−7f′(u4)=3(−0.31603…)2+8(−0.31603…)−2=−4.22862…u5=−0.31603…
Δu5=∣−0.31603…−(−0.31603…)∣=3.13479E−8Δu5=3.13479E−8
u≈−0.31603…
긴 나눗셈 적용:u+0.31603…u3+4u2−2u−1=u2+3.68396…u−3.16424…
u2+3.68396…u−3.16424…≈0
다음을 위한 하나의 솔루션 찾기 u2+3.68396…u−3.16424…=0 뉴턴-랩슨을 이용하여:u≈0.71870…
u2+3.68396…u−3.16424…=0
뉴턴-랩슨 근사 정의
f(u)=u2+3.68396…u−3.16424…
f′(u)찾다 :2u+3.68396…
dud(u2+3.68396…u−3.16424…)
합계/차이 규칙 적용: (f±g)′=f′±g′=dud(u2)+dud(3.68396…u)−dud(3.16424…)
dud(u2)=2u
dud(u2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=2u2−1
단순화=2u
dud(3.68396…u)=3.68396…
dud(3.68396…u)
정수를 빼라: (a⋅f)′=a⋅f′=3.68396…dudu
공통 도함수 적용: dudu=1=3.68396…⋅1
단순화=3.68396…
dud(3.16424…)=0
dud(3.16424…)
상수의 도함수: dxd(a)=0=0
=2u+3.68396…−0
단순화=2u+3.68396…
렛 u0=1계산하다 un+1 까지 Δun+1<0.000001
u1=0.73263…:Δu1=0.26736…
f(u0)=12+3.68396…⋅1−3.16424…=1.51972…f′(u0)=2⋅1+3.68396…=5.68396…u1=0.73263…
Δu1=∣0.73263…−1∣=0.26736…Δu1=0.26736…
u2=0.71874…:Δu2=0.01388…
f(u1)=0.73263…2+3.68396…⋅0.73263…−3.16424…=0.07148…f′(u1)=2⋅0.73263…+3.68396…=5.14922…u2=0.71874…
Δu2=∣0.71874…−0.73263…∣=0.01388…Δu2=0.01388…
u3=0.71870…:Δu3=0.00003…
f(u2)=0.71874…2+3.68396…⋅0.71874…−3.16424…=0.00019…f′(u2)=2⋅0.71874…+3.68396…=5.12146…u3=0.71870…
Δu3=∣0.71870…−0.71874…∣=0.00003…Δu3=0.00003…
u4=0.71870…:Δu4=2.76537E−10
f(u3)=0.71870…2+3.68396…⋅0.71870…−3.16424…=1.41625E−9f′(u3)=2⋅0.71870…+3.68396…=5.12138…u4=0.71870…
Δu4=∣0.71870…−0.71870…∣=2.76537E−10Δu4=2.76537E−10
u≈0.71870…
긴 나눗셈 적용:u−0.71870…u2+3.68396…u−3.16424…=u+4.40267…
u+4.40267…≈0
u≈−4.40267…
해결책은u≈−0.31603…,u≈0.71870…,u≈−4.40267…
뒤로 대체 u=sin(x)sin(x)≈−0.31603…,sin(x)≈0.71870…,sin(x)≈−4.40267…
sin(x)≈−0.31603…,sin(x)≈0.71870…,sin(x)≈−4.40267…
sin(x)=−0.31603…:x=arcsin(−0.31603…)+2πn,x=π+arcsin(0.31603…)+2πn
sin(x)=−0.31603…
트리거 역속성 적용
sin(x)=−0.31603…
일반 솔루션 sin(x)=−0.31603…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(−0.31603…)+2πn,x=π+arcsin(0.31603…)+2πn
x=arcsin(−0.31603…)+2πn,x=π+arcsin(0.31603…)+2πn
sin(x)=0.71870…:x=arcsin(0.71870…)+2πn,x=π−arcsin(0.71870…)+2πn
sin(x)=0.71870…
트리거 역속성 적용
sin(x)=0.71870…
일반 솔루션 sin(x)=0.71870…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(0.71870…)+2πn,x=π−arcsin(0.71870…)+2πn
x=arcsin(0.71870…)+2πn,x=π−arcsin(0.71870…)+2πn
sin(x)=−4.40267…:해결책 없음
sin(x)=−4.40267…
−1≤sin(x)≤1해결책없음
모든 솔루션 결합x=arcsin(−0.31603…)+2πn,x=π+arcsin(0.31603…)+2πn,x=arcsin(0.71870…)+2πn,x=π−arcsin(0.71870…)+2πn
해를 10진수 형식으로 표시x=−0.32154…+2πn,x=π+0.32154…+2πn,x=0.80194…+2πn,x=π−0.80194…+2πn