해법
12=3sec(θ)+5csc(θ)
해법
θ=π−0.33570…+2πn,θ=1.07609…+2πn,θ=0.65383…+2πn,θ=−1.39422…+2πn
+1
도
θ=160.76534…∘+360∘n,θ=61.65590…∘+360∘n,θ=37.46199…∘+360∘n,θ=−79.88324…∘+360∘n솔루션 단계
12=3sec(θ)+5csc(θ)
빼다 5csc(θ) 양쪽에서3sec(θ)=12−5csc(θ)
양쪽을 제곱(3sec(θ))2=(12−5csc(θ))2
빼다 (12−5csc(θ))2 양쪽에서9sec2(θ)−144+120csc(θ)−25csc2(θ)=0
죄로 표현하라, 왜냐하면
−144+120csc(θ)−25csc2(θ)+9sec2(θ)
기본 삼각형 항등식 사용: csc(x)=sin(x)1=−144+120⋅sin(θ)1−25(sin(θ)1)2+9sec2(θ)
기본 삼각형 항등식 사용: sec(x)=cos(x)1=−144+120⋅sin(θ)1−25(sin(θ)1)2+9(cos(θ)1)2
−144+120⋅sin(θ)1−25(sin(θ)1)2+9(cos(θ)1)2단순화하세요:cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)
−144+120⋅sin(θ)1−25(sin(θ)1)2+9(cos(θ)1)2
120⋅sin(θ)1=sin(θ)120
120⋅sin(θ)1
다중 분수: a⋅cb=ca⋅b=sin(θ)1⋅120
숫자를 곱하시오: 1⋅120=120=sin(θ)120
25(sin(θ)1)2=sin2(θ)25
25(sin(θ)1)2
(sin(θ)1)2=sin2(θ)1
(sin(θ)1)2
지수 규칙 적용: (ba)c=bcac=sin2(θ)12
규칙 적용 1a=112=1=sin2(θ)1
=25⋅sin2(θ)1
다중 분수: a⋅cb=ca⋅b=sin2(θ)1⋅25
숫자를 곱하시오: 1⋅25=25=sin2(θ)25
9(cos(θ)1)2=cos2(θ)9
9(cos(θ)1)2
(cos(θ)1)2=cos2(θ)1
(cos(θ)1)2
지수 규칙 적용: (ba)c=bcac=cos2(θ)12
규칙 적용 1a=112=1=cos2(θ)1
=9⋅cos2(θ)1
다중 분수: a⋅cb=ca⋅b=cos2(θ)1⋅9
숫자를 곱하시오: 1⋅9=9=cos2(θ)9
=−144+sin(θ)120−sin2(θ)25+cos2(θ)9
요소를 분수로 변환: 144=cos2(θ)144cos2(θ)=−cos2(θ)144cos2(θ)+sin(θ)120−sin2(θ)25+cos2(θ)9
cos2(θ),sin(θ),sin2(θ),cos2(θ) 의 최소 공배수:cos2(θ)sin2(θ)
cos2(θ),sin(θ),sin2(θ),cos2(θ)
최저공통승수 (LCM)
인수식 중 하나 이상에 나타나는 요인으로 구성된 식을 계산합니다=cos2(θ)sin2(θ)
LCM을 기준으로 분수 조정
각 분자를 곱하는 데 필요한 동일한 양으로 곱하시오
해당 분모를 LCM으로 변환합니다 cos2(θ)sin2(θ)
위해서 cos2(θ)144cos2(θ):분모와 분자를 곱하다 sin2(θ)cos2(θ)144cos2(θ)=cos2(θ)sin2(θ)144cos2(θ)sin2(θ)
위해서 sin(θ)120:분모와 분자를 곱하다 cos2(θ)sin(θ)sin(θ)120=sin(θ)cos2(θ)sin(θ)120cos2(θ)sin(θ)=cos2(θ)sin2(θ)120cos2(θ)sin(θ)
위해서 sin2(θ)25:분모와 분자를 곱하다 cos2(θ)sin2(θ)25=sin2(θ)cos2(θ)25cos2(θ)
위해서 cos2(θ)9:분모와 분자를 곱하다 sin2(θ)cos2(θ)9=cos2(θ)sin2(θ)9sin2(θ)
=−cos2(θ)sin2(θ)144cos2(θ)sin2(θ)+cos2(θ)sin2(θ)120cos2(θ)sin(θ)−sin2(θ)cos2(θ)25cos2(θ)+cos2(θ)sin2(θ)9sin2(θ)
분모가 같기 때문에, 분수를 합친다: ca±cb=ca±b=cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)
=cos2(θ)sin2(θ)−144cos2(θ)sin2(θ)+120cos2(θ)sin(θ)−25cos2(θ)+9sin2(θ)
cos2(θ)sin2(θ)−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)=0
g(x)f(x)=0⇒f(x)=0−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)=0
삼각성을 사용하여 다시 쓰기
−25cos2(θ)+9sin2(θ)+120cos2(θ)sin(θ)−144cos2(θ)sin2(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
−25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)간소화하다 :−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25(1−sin2(θ))+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
=−25(1−sin2(θ))+9sin2(θ)+120sin(θ)(1−sin2(θ))−144sin2(θ)(1−sin2(θ))
−25(1−sin2(θ))확대한다:−25+25sin2(θ)
−25(1−sin2(θ))
분배 법칙 적용: a(b−c)=ab−aca=−25,b=1,c=sin2(θ)=−25⋅1−(−25)sin2(θ)
마이너스 플러스 규칙 적용−(−a)=a=−25⋅1+25sin2(θ)
숫자를 곱하시오: 25⋅1=25=−25+25sin2(θ)
=−25+25sin2(θ)+9sin2(θ)+120(1−sin2(θ))sin(θ)−144(1−sin2(θ))sin2(θ)
120sin(θ)(1−sin2(θ))확대한다:120sin(θ)−120sin3(θ)
120sin(θ)(1−sin2(θ))
분배 법칙 적용: a(b−c)=ab−aca=120sin(θ),b=1,c=sin2(θ)=120sin(θ)⋅1−120sin(θ)sin2(θ)
=120⋅1⋅sin(θ)−120sin2(θ)sin(θ)
120⋅1⋅sin(θ)−120sin2(θ)sin(θ)단순화하세요:120sin(θ)−120sin3(θ)
120⋅1⋅sin(θ)−120sin2(θ)sin(θ)
120⋅1⋅sin(θ)=120sin(θ)
120⋅1⋅sin(θ)
숫자를 곱하시오: 120⋅1=120=120sin(θ)
120sin2(θ)sin(θ)=120sin3(θ)
120sin2(θ)sin(θ)
지수 규칙 적용: ab⋅ac=ab+csin2(θ)sin(θ)=sin2+1(θ)=120sin2+1(θ)
숫자 추가: 2+1=3=120sin3(θ)
=120sin(θ)−120sin3(θ)
=120sin(θ)−120sin3(θ)
=−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144(1−sin2(θ))sin2(θ)
−144sin2(θ)(1−sin2(θ))확대한다:−144sin2(θ)+144sin4(θ)
−144sin2(θ)(1−sin2(θ))
분배 법칙 적용: a(b−c)=ab−aca=−144sin2(θ),b=1,c=sin2(θ)=−144sin2(θ)⋅1−(−144sin2(θ))sin2(θ)
마이너스 플러스 규칙 적용−(−a)=a=−144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ)
−144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ)단순화하세요:−144sin2(θ)+144sin4(θ)
−144⋅1⋅sin2(θ)+144sin2(θ)sin2(θ)
144⋅1⋅sin2(θ)=144sin2(θ)
144⋅1⋅sin2(θ)
숫자를 곱하시오: 144⋅1=144=144sin2(θ)
144sin2(θ)sin2(θ)=144sin4(θ)
144sin2(θ)sin2(θ)
지수 규칙 적용: ab⋅ac=ab+csin2(θ)sin2(θ)=sin2+2(θ)=144sin2+2(θ)
숫자 추가: 2+2=4=144sin4(θ)
=−144sin2(θ)+144sin4(θ)
=−144sin2(θ)+144sin4(θ)
=−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)
−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)단순화하세요:−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25+25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)
집단적 용어=25sin2(θ)+9sin2(θ)+120sin(θ)−120sin3(θ)−144sin2(θ)+144sin4(θ)−25
유사 요소 추가: 25sin2(θ)+9sin2(θ)−144sin2(θ)=−110sin2(θ)=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
=−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)−25
−25−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)=0
대체로 해결
−25−110sin2(θ)+120sin(θ)−120sin3(θ)+144sin4(θ)=0
하게: sin(θ)=u−25−110u2+120u−120u3+144u4=0
−25−110u2+120u−120u3+144u4=0:u≈0.32943…,u≈0.88011…,u≈0.60823…,u≈−0.98445…
−25−110u2+120u−120u3+144u4=0
표준 양식으로 작성 anxn+…+a1x+a0=0144u4−120u3−110u2+120u−25=0
다음을 위한 하나의 솔루션 찾기 144u4−120u3−110u2+120u−25=0 뉴턴-랩슨을 이용하여:u≈0.32943…
144u4−120u3−110u2+120u−25=0
뉴턴-랩슨 근사 정의
f(u)=144u4−120u3−110u2+120u−25
f′(u)찾다 :576u3−360u2−220u+120
dud(144u4−120u3−110u2+120u−25)
합계/차이 규칙 적용: (f±g)′=f′±g′=dud(144u4)−dud(120u3)−dud(110u2)+dud(120u)−dud(25)
dud(144u4)=576u3
dud(144u4)
정수를 빼라: (a⋅f)′=a⋅f′=144dud(u4)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=144⋅4u4−1
단순화=576u3
dud(120u3)=360u2
dud(120u3)
정수를 빼라: (a⋅f)′=a⋅f′=120dud(u3)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=120⋅3u3−1
단순화=360u2
dud(110u2)=220u
dud(110u2)
정수를 빼라: (a⋅f)′=a⋅f′=110dud(u2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=110⋅2u2−1
단순화=220u
dud(120u)=120
dud(120u)
정수를 빼라: (a⋅f)′=a⋅f′=120dudu
공통 도함수 적용: dudu=1=120⋅1
단순화=120
dud(25)=0
dud(25)
상수의 도함수: dxd(a)=0=0
=576u3−360u2−220u+120−0
단순화=576u3−360u2−220u+120
렛 u0=0계산하다 un+1 까지 Δun+1<0.000001
u1=0.20833…:Δu1=0.20833…
f(u0)=144⋅04−120⋅03−110⋅02+120⋅0−25=−25f′(u0)=576⋅03−360⋅02−220⋅0+120=120u1=0.20833…
Δu1=∣0.20833…−0∣=0.20833…Δu1=0.20833…
u2=0.29598…:Δu2=0.08765…
f(u1)=144⋅0.20833…4−120⋅0.20833…3−110⋅0.20833…2+120⋅0.20833…−25=−5.58810…f′(u1)=576⋅0.20833…3−360⋅0.20833…2−220⋅0.20833…+120=63.75u2=0.29598…
Δu2=∣0.29598…−0.20833…∣=0.08765…Δu2=0.08765…
u3=0.32537…:Δu3=0.02938…
f(u2)=144⋅0.29598…4−120⋅0.29598…3−110⋅0.29598…2+120⋅0.29598…−25=−1.12484…f′(u2)=576⋅0.29598…3−360⋅0.29598…2−220⋅0.29598…+120=38.27925…u3=0.32537…
Δu3=∣0.32537…−0.29598…∣=0.02938…Δu3=0.02938…
u4=0.32936…:Δu4=0.00398…
f(u3)=144⋅0.32537…4−120⋅0.32537…3−110⋅0.32537…2+120⋅0.32537…−25=−0.12024…f′(u3)=576⋅0.32537…3−360⋅0.32537…2−220⋅0.32537…+120=30.14620…u4=0.32936…
Δu4=∣0.32936…−0.32537…∣=0.00398…Δu4=0.00398…
u5=0.32943…:Δu5=0.00007…
f(u4)=144⋅0.32936…4−120⋅0.32936…3−110⋅0.32936…2+120⋅0.32936…−25=−0.00215…f′(u4)=576⋅0.32936…3−360⋅0.32936…2−220⋅0.32936…+120=29.06722…u5=0.32943…
Δu5=∣0.32943…−0.32936…∣=0.00007…Δu5=0.00007…
u6=0.32943…:Δu6=2.54922E−8
f(u5)=144⋅0.32943…4−120⋅0.32943…3−110⋅0.32943…2+120⋅0.32943…−25=−7.40478E−7f′(u5)=576⋅0.32943…3−360⋅0.32943…2−220⋅0.32943…+120=29.04723…u6=0.32943…
Δu6=∣0.32943…−0.32943…∣=2.54922E−8Δu6=2.54922E−8
u≈0.32943…
긴 나눗셈 적용:u−0.32943…144u4−120u3−110u2+120u−25=144u3−72.56094…u2−133.90432…u+75.88684…
144u3−72.56094…u2−133.90432…u+75.88684…≈0
다음을 위한 하나의 솔루션 찾기 144u3−72.56094…u2−133.90432…u+75.88684…=0 뉴턴-랩슨을 이용하여:u≈0.88011…
144u3−72.56094…u2−133.90432…u+75.88684…=0
뉴턴-랩슨 근사 정의
f(u)=144u3−72.56094…u2−133.90432…u+75.88684…
f′(u)찾다 :432u2−145.12189…u−133.90432…
dud(144u3−72.56094…u2−133.90432…u+75.88684…)
합계/차이 규칙 적용: (f±g)′=f′±g′=dud(144u3)−dud(72.56094…u2)−dud(133.90432…u)+dud(75.88684…)
dud(144u3)=432u2
dud(144u3)
정수를 빼라: (a⋅f)′=a⋅f′=144dud(u3)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=144⋅3u3−1
단순화=432u2
dud(72.56094…u2)=145.12189…u
dud(72.56094…u2)
정수를 빼라: (a⋅f)′=a⋅f′=72.56094…dud(u2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=72.56094…⋅2u2−1
단순화=145.12189…u
dud(133.90432…u)=133.90432…
dud(133.90432…u)
정수를 빼라: (a⋅f)′=a⋅f′=133.90432…dudu
공통 도함수 적용: dudu=1=133.90432…⋅1
단순화=133.90432…
dud(75.88684…)=0
dud(75.88684…)
상수의 도함수: dxd(a)=0=0
=432u2−145.12189…u−133.90432…+0
단순화=432u2−145.12189…u−133.90432…
렛 u0=1계산하다 un+1 까지 Δun+1<0.000001
u1=0.91226…:Δu1=0.08773…
f(u0)=144⋅13−72.56094…⋅12−133.90432…⋅1+75.88684…=13.42157…f′(u0)=432⋅12−145.12189…⋅1−133.90432…=152.97377…u1=0.91226…
Δu1=∣0.91226…−1∣=0.08773…Δu1=0.08773…
u2=0.88362…:Δu2=0.02863…
f(u1)=144⋅0.91226…3−72.56094…⋅0.91226…2−133.90432…⋅0.91226…+75.88684…=2.66967…f′(u1)=432⋅0.91226…2−145.12189…⋅0.91226…−133.90432…=93.22653…u2=0.88362…
Δu2=∣0.88362…−0.91226…∣=0.02863…Δu2=0.02863…
u3=0.88016…:Δu3=0.00346…
f(u2)=144⋅0.88362…3−72.56094…⋅0.88362…2−133.90432…⋅0.88362…+75.88684…=0.26029…f′(u2)=432⋅0.88362…2−145.12189…⋅0.88362…−133.90432…=75.16549…u3=0.88016…
Δu3=∣0.88016…−0.88362…∣=0.00346…Δu3=0.00346…
u4=0.88011…:Δu4=0.00005…
f(u3)=144⋅0.88016…3−72.56094…⋅0.88016…2−133.90432…⋅0.88016…+75.88684…=0.00370…f′(u3)=432⋅0.88016…2−145.12189…⋅0.88016…−133.90432…=73.02944…u4=0.88011…
Δu4=∣0.88011…−0.88016…∣=0.00005…Δu4=0.00005…
u5=0.88011…:Δu5=1.08272E−8
f(u4)=144⋅0.88011…3−72.56094…⋅0.88011…2−133.90432…⋅0.88011…+75.88684…=7.90368E−7f′(u4)=432⋅0.88011…2−145.12189…⋅0.88011…−133.90432…=72.99825…u5=0.88011…
Δu5=∣0.88011…−0.88011…∣=1.08272E−8Δu5=1.08272E−8
u≈0.88011…
긴 나눗셈 적용:u−0.88011…144u3−72.56094…u2−133.90432…u+75.88684…=144u2+54.17521…u−86.22405…
144u2+54.17521…u−86.22405…≈0
다음을 위한 하나의 솔루션 찾기 144u2+54.17521…u−86.22405…=0 뉴턴-랩슨을 이용하여:u≈0.60823…
144u2+54.17521…u−86.22405…=0
뉴턴-랩슨 근사 정의
f(u)=144u2+54.17521…u−86.22405…
f′(u)찾다 :288u+54.17521…
dud(144u2+54.17521…u−86.22405…)
합계/차이 규칙 적용: (f±g)′=f′±g′=dud(144u2)+dud(54.17521…u)−dud(86.22405…)
dud(144u2)=288u
dud(144u2)
정수를 빼라: (a⋅f)′=a⋅f′=144dud(u2)
전원 규칙을 적용합니다: dxd(xa)=a⋅xa−1=144⋅2u2−1
단순화=288u
dud(54.17521…u)=54.17521…
dud(54.17521…u)
정수를 빼라: (a⋅f)′=a⋅f′=54.17521…dudu
공통 도함수 적용: dudu=1=54.17521…⋅1
단순화=54.17521…
dud(86.22405…)=0
dud(86.22405…)
상수의 도함수: dxd(a)=0=0
=288u+54.17521…−0
단순화=288u+54.17521…
렛 u0=2계산하다 un+1 까지 Δun+1<0.000001
u1=1.05085…:Δu1=0.94914…
f(u0)=144⋅22+54.17521…⋅2−86.22405…=598.12636…f′(u0)=288⋅2+54.17521…=630.17521…u1=1.05085…
Δu1=∣1.05085…−2∣=0.94914…Δu1=0.94914…
u2=0.68729…:Δu2=0.36355…
f(u1)=144⋅1.05085…2+54.17521…⋅1.05085…−86.22405…=129.72561…f′(u1)=288⋅1.05085…+54.17521…=356.82203…u2=0.68729…
Δu2=∣0.68729…−1.05085…∣=0.36355…Δu2=0.36355…
u3=0.61180…:Δu3=0.07549…
f(u2)=144⋅0.68729…2+54.17521…⋅0.68729…−86.22405…=19.03314…f′(u2)=288⋅0.68729…+54.17521…=252.11724…u3=0.61180…
Δu3=∣0.61180…−0.68729…∣=0.07549…Δu3=0.07549…
u4=0.60824…:Δu4=0.00356…
f(u3)=144⋅0.61180…2+54.17521…⋅0.61180…−86.22405…=0.82068…f′(u3)=288⋅0.61180…+54.17521…=230.37518…u4=0.60824…
Δu4=∣0.60824…−0.61180…∣=0.00356…Δu4=0.00356…
u5=0.60823…:Δu5=7.96803E−6
f(u4)=144⋅0.60824…2+54.17521…⋅0.60824…−86.22405…=0.00182…f′(u4)=288⋅0.60824…+54.17521…=229.34921…u5=0.60823…
Δu5=∣0.60823…−0.60824…∣=7.96803E−6Δu5=7.96803E−6
u6=0.60823…:Δu6=3.98632E−11
f(u5)=144⋅0.60823…2+54.17521…⋅0.60823…−86.22405…=9.1425E−9f′(u5)=288⋅0.60823…+54.17521…=229.34692…u6=0.60823…
Δu6=∣0.60823…−0.60823…∣=3.98632E−11Δu6=3.98632E−11
u≈0.60823…
긴 나눗셈 적용:u−0.60823…144u2+54.17521…u−86.22405…=144u+141.76106…
144u+141.76106…≈0
u≈−0.98445…
해결책은u≈0.32943…,u≈0.88011…,u≈0.60823…,u≈−0.98445…
뒤로 대체 u=sin(θ)sin(θ)≈0.32943…,sin(θ)≈0.88011…,sin(θ)≈0.60823…,sin(θ)≈−0.98445…
sin(θ)≈0.32943…,sin(θ)≈0.88011…,sin(θ)≈0.60823…,sin(θ)≈−0.98445…
sin(θ)=0.32943…:θ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
sin(θ)=0.32943…
트리거 역속성 적용
sin(θ)=0.32943…
일반 솔루션 sin(θ)=0.32943…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
θ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn
sin(θ)=0.88011…:θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
sin(θ)=0.88011…
트리거 역속성 적용
sin(θ)=0.88011…
일반 솔루션 sin(θ)=0.88011…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn
sin(θ)=0.60823…:θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
sin(θ)=0.60823…
트리거 역속성 적용
sin(θ)=0.60823…
일반 솔루션 sin(θ)=0.60823…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn
sin(θ)=−0.98445…:θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
sin(θ)=−0.98445…
트리거 역속성 적용
sin(θ)=−0.98445…
일반 솔루션 sin(θ)=−0.98445…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnθ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
모든 솔루션 결합θ=arcsin(0.32943…)+2πn,θ=π−arcsin(0.32943…)+2πn,θ=arcsin(0.88011…)+2πn,θ=π−arcsin(0.88011…)+2πn,θ=arcsin(0.60823…)+2πn,θ=π−arcsin(0.60823…)+2πn,θ=arcsin(−0.98445…)+2πn,θ=π+arcsin(0.98445…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 3sec(θ)+5csc(θ)=12
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arcsin(0.32943…)+2πn:거짓
arcsin(0.32943…)+2πn
n=1끼우다 arcsin(0.32943…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=arcsin(0.32943…)+2π13sec(arcsin(0.32943…)+2π1)+5csc(arcsin(0.32943…)+2π1)=12
다듬다18.35473…=12
⇒거짓
솔루션 확인 π−arcsin(0.32943…)+2πn:참
π−arcsin(0.32943…)+2πn
n=1끼우다 π−arcsin(0.32943…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=π−arcsin(0.32943…)+2π13sec(π−arcsin(0.32943…)+2π1)+5csc(π−arcsin(0.32943…)+2π1)=12
다듬다12=12
⇒참
솔루션 확인 arcsin(0.88011…)+2πn:참
arcsin(0.88011…)+2πn
n=1끼우다 arcsin(0.88011…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=arcsin(0.88011…)+2π13sec(arcsin(0.88011…)+2π1)+5csc(arcsin(0.88011…)+2π1)=12
다듬다12=12
⇒참
솔루션 확인 π−arcsin(0.88011…)+2πn:거짓
π−arcsin(0.88011…)+2πn
n=1끼우다 π−arcsin(0.88011…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=π−arcsin(0.88011…)+2π13sec(π−arcsin(0.88011…)+2π1)+5csc(π−arcsin(0.88011…)+2π1)=12
다듬다−0.63781…=12
⇒거짓
솔루션 확인 arcsin(0.60823…)+2πn:참
arcsin(0.60823…)+2πn
n=1끼우다 arcsin(0.60823…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=arcsin(0.60823…)+2π13sec(arcsin(0.60823…)+2π1)+5csc(arcsin(0.60823…)+2π1)=12
다듬다12=12
⇒참
솔루션 확인 π−arcsin(0.60823…)+2πn:거짓
π−arcsin(0.60823…)+2πn
n=1끼우다 π−arcsin(0.60823…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=π−arcsin(0.60823…)+2π13sec(π−arcsin(0.60823…)+2π1)+5csc(π−arcsin(0.60823…)+2π1)=12
다듬다4.44101…=12
⇒거짓
솔루션 확인 arcsin(−0.98445…)+2πn:참
arcsin(−0.98445…)+2πn
n=1끼우다 arcsin(−0.98445…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=arcsin(−0.98445…)+2π13sec(arcsin(−0.98445…)+2π1)+5csc(arcsin(−0.98445…)+2π1)=12
다듬다12=12
⇒참
솔루션 확인 π+arcsin(0.98445…)+2πn:거짓
π+arcsin(0.98445…)+2πn
n=1끼우다 π+arcsin(0.98445…)+2π1
3sec(θ)+5csc(θ)=12 위한 {\ quad}끼우다{\ quad} θ=π+arcsin(0.98445…)+2π13sec(π+arcsin(0.98445…)+2π1)+5csc(π+arcsin(0.98445…)+2π1)=12
다듬다−22.15793…=12
⇒거짓
θ=π−arcsin(0.32943…)+2πn,θ=arcsin(0.88011…)+2πn,θ=arcsin(0.60823…)+2πn,θ=arcsin(−0.98445…)+2πn
해를 10진수 형식으로 표시θ=π−0.33570…+2πn,θ=1.07609…+2πn,θ=0.65383…+2πn,θ=−1.39422…+2πn