해법
sin(θ)−0.1cos(θ)=9.88.87
해법
θ=2.12008…+2πn,θ=1.22084…+2πn
+1
도
θ=121.47220…∘+360∘n,θ=69.94898…∘+360∘n솔루션 단계
sin(θ)−0.1cos(θ)=9.88.87
더하다 0.1cos(θ) 양쪽으로sin(θ)=0.90510…+0.1cos(θ)
양쪽을 제곱sin2(θ)=(0.90510…+0.1cos(θ))2
빼다 (0.90510…+0.1cos(θ))2 양쪽에서sin2(θ)−0.81920…−0.18102…cos(θ)−0.01cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−0.81920…+sin2(θ)−0.01cos2(θ)−0.18102…cos(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.81920…+1−cos2(θ)−0.01cos2(θ)−0.18102…cos(θ)
−0.81920…+1−cos2(θ)−0.01cos2(θ)−0.18102…cos(θ)간소화하다 :−1.01cos2(θ)−0.18102…cos(θ)+0.18079…
−0.81920…+1−cos2(θ)−0.01cos2(θ)−0.18102…cos(θ)
유사 요소 추가: −cos2(θ)−0.01cos2(θ)=−1.01cos2(θ)=−0.81920…+1−1.01cos2(θ)−0.18102…cos(θ)
숫자 더하기/ 빼기: −0.81920…+1=0.18079…=−1.01cos2(θ)−0.18102…cos(θ)+0.18079…
=−1.01cos2(θ)−0.18102…cos(θ)+0.18079…
0.18079…−0.18102…cos(θ)−1.01cos2(θ)=0
대체로 해결
0.18079…−0.18102…cos(θ)−1.01cos2(θ)=0
하게: cos(θ)=u0.18079…−0.18102…u−1.01u2=0
0.18079…−0.18102…u−1.01u2=0:u=−2.020.18102…+0.76316…,u=2.020.76316…−0.18102…
0.18079…−0.18102…u−1.01u2=0
표준 양식으로 작성 ax2+bx+c=0−1.01u2−0.18102…u+0.18079…=0
쿼드 공식으로 해결
−1.01u2−0.18102…u+0.18079…=0
4차 방정식 공식:
위해서 a=−1.01,b=−0.18102…,c=0.18079…u1,2=2(−1.01)−(−0.18102…)±(−0.18102…)2−4(−1.01)⋅0.18079…
u1,2=2(−1.01)−(−0.18102…)±(−0.18102…)2−4(−1.01)⋅0.18079…
(−0.18102…)2−4(−1.01)⋅0.18079…=0.76316…
(−0.18102…)2−4(−1.01)⋅0.18079…
규칙 적용 −(−a)=a=(−0.18102…)2+4⋅1.01⋅0.18079…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.18102…)2=0.18102…2=0.18102…2+4⋅0.18079…⋅1.01
숫자를 곱하시오: 4⋅1.01⋅0.18079…=0.73039…=0.18102…2+0.73039…
0.18102…2=0.03276…=0.03276…+0.73039…
숫자 추가: 0.03276…+0.73039…=0.76316…=0.76316…
u1,2=2(−1.01)−(−0.18102…)±0.76316…
솔루션 분리u1=2(−1.01)−(−0.18102…)+0.76316…,u2=2(−1.01)−(−0.18102…)−0.76316…
u=2(−1.01)−(−0.18102…)+0.76316…:−2.020.18102…+0.76316…
2(−1.01)−(−0.18102…)+0.76316…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.18102…+0.76316…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.18102…+0.76316…
분수 규칙 적용: −ba=−ba=−2.020.18102…+0.76316…
u=2(−1.01)−(−0.18102…)−0.76316…:2.020.76316…−0.18102…
2(−1.01)−(−0.18102…)−0.76316…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.18102…−0.76316…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.18102…−0.76316…
분수 규칙 적용: −b−a=ba0.18102…−0.76316…=−(0.76316…−0.18102…)=2.020.76316…−0.18102…
2차 방정식의 해는 다음과 같다:u=−2.020.18102…+0.76316…,u=2.020.76316…−0.18102…
뒤로 대체 u=cos(θ)cos(θ)=−2.020.18102…+0.76316…,cos(θ)=2.020.76316…−0.18102…
cos(θ)=−2.020.18102…+0.76316…,cos(θ)=2.020.76316…−0.18102…
cos(θ)=−2.020.18102…+0.76316…:θ=arccos(−2.020.18102…+0.76316…)+2πn,θ=−arccos(−2.020.18102…+0.76316…)+2πn
cos(θ)=−2.020.18102…+0.76316…
트리거 역속성 적용
cos(θ)=−2.020.18102…+0.76316…
일반 솔루션 cos(θ)=−2.020.18102…+0.76316…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.020.18102…+0.76316…)+2πn,θ=−arccos(−2.020.18102…+0.76316…)+2πn
θ=arccos(−2.020.18102…+0.76316…)+2πn,θ=−arccos(−2.020.18102…+0.76316…)+2πn
cos(θ)=2.020.76316…−0.18102…:θ=arccos(2.020.76316…−0.18102…)+2πn,θ=2π−arccos(2.020.76316…−0.18102…)+2πn
cos(θ)=2.020.76316…−0.18102…
트리거 역속성 적용
cos(θ)=2.020.76316…−0.18102…
일반 솔루션 cos(θ)=2.020.76316…−0.18102…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.020.76316…−0.18102…)+2πn,θ=2π−arccos(2.020.76316…−0.18102…)+2πn
θ=arccos(2.020.76316…−0.18102…)+2πn,θ=2π−arccos(2.020.76316…−0.18102…)+2πn
모든 솔루션 결합θ=arccos(−2.020.18102…+0.76316…)+2πn,θ=−arccos(−2.020.18102…+0.76316…)+2πn,θ=arccos(2.020.76316…−0.18102…)+2πn,θ=2π−arccos(2.020.76316…−0.18102…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(θ)−0.1cos(θ)=9.88.87
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.020.18102…+0.76316…)+2πn:참
arccos(−2.020.18102…+0.76316…)+2πn
n=1끼우다 arccos(−2.020.18102…+0.76316…)+2π1
sin(θ)−0.1cos(θ)=9.88.87 위한 {\ quad}끼우다{\ quad} θ=arccos(−2.020.18102…+0.76316…)+2π1sin(arccos(−2.020.18102…+0.76316…)+2π1)−0.1cos(arccos(−2.020.18102…+0.76316…)+2π1)=9.88.87
다듬다0.90510…=0.90510…
⇒참
솔루션 확인 −arccos(−2.020.18102…+0.76316…)+2πn:거짓
−arccos(−2.020.18102…+0.76316…)+2πn
n=1끼우다 −arccos(−2.020.18102…+0.76316…)+2π1
sin(θ)−0.1cos(θ)=9.88.87 위한 {\ quad}끼우다{\ quad} θ=−arccos(−2.020.18102…+0.76316…)+2π1sin(−arccos(−2.020.18102…+0.76316…)+2π1)−0.1cos(−arccos(−2.020.18102…+0.76316…)+2π1)=9.88.87
다듬다−0.80068…=0.90510…
⇒거짓
솔루션 확인 arccos(2.020.76316…−0.18102…)+2πn:참
arccos(2.020.76316…−0.18102…)+2πn
n=1끼우다 arccos(2.020.76316…−0.18102…)+2π1
sin(θ)−0.1cos(θ)=9.88.87 위한 {\ quad}끼우다{\ quad} θ=arccos(2.020.76316…−0.18102…)+2π1sin(arccos(2.020.76316…−0.18102…)+2π1)−0.1cos(arccos(2.020.76316…−0.18102…)+2π1)=9.88.87
다듬다0.90510…=0.90510…
⇒참
솔루션 확인 2π−arccos(2.020.76316…−0.18102…)+2πn:거짓
2π−arccos(2.020.76316…−0.18102…)+2πn
n=1끼우다 2π−arccos(2.020.76316…−0.18102…)+2π1
sin(θ)−0.1cos(θ)=9.88.87 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(2.020.76316…−0.18102…)+2π1sin(2π−arccos(2.020.76316…−0.18102…)+2π1)−0.1cos(2π−arccos(2.020.76316…−0.18102…)+2π1)=9.88.87
다듬다−0.97367…=0.90510…
⇒거짓
θ=arccos(−2.020.18102…+0.76316…)+2πn,θ=arccos(2.020.76316…−0.18102…)+2πn
해를 10진수 형식으로 표시θ=2.12008…+2πn,θ=1.22084…+2πn