해법
sin(A)−0.1⋅cos(A)=9.86.94
해법
A=2.45933…+2πn,A=0.88159…+2πn
+1
도
A=140.90941…∘+360∘n,A=50.51177…∘+360∘n솔루션 단계
sin(A)−0.1cos(A)=9.86.94
더하다 0.1cos(A) 양쪽으로sin(A)=0.70816…+0.1cos(A)
양쪽을 제곱sin2(A)=(0.70816…+0.1cos(A))2
빼다 (0.70816…+0.1cos(A))2 양쪽에서sin2(A)−0.50149…−0.14163…cos(A)−0.01cos2(A)=0
삼각성을 사용하여 다시 쓰기
−0.50149…+sin2(A)−0.01cos2(A)−0.14163…cos(A)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.50149…+1−cos2(A)−0.01cos2(A)−0.14163…cos(A)
−0.50149…+1−cos2(A)−0.01cos2(A)−0.14163…cos(A)간소화하다 :−1.01cos2(A)−0.14163…cos(A)+0.49850…
−0.50149…+1−cos2(A)−0.01cos2(A)−0.14163…cos(A)
유사 요소 추가: −cos2(A)−0.01cos2(A)=−1.01cos2(A)=−0.50149…+1−1.01cos2(A)−0.14163…cos(A)
숫자 더하기/ 빼기: −0.50149…+1=0.49850…=−1.01cos2(A)−0.14163…cos(A)+0.49850…
=−1.01cos2(A)−0.14163…cos(A)+0.49850…
0.49850…−0.14163…cos(A)−1.01cos2(A)=0
대체로 해결
0.49850…−0.14163…cos(A)−1.01cos2(A)=0
하게: cos(A)=u0.49850…−0.14163…u−1.01u2=0
0.49850…−0.14163…u−1.01u2=0:u=−2.020.14163…+2.03401…,u=2.022.03401…−0.14163…
0.49850…−0.14163…u−1.01u2=0
표준 양식으로 작성 ax2+bx+c=0−1.01u2−0.14163…u+0.49850…=0
쿼드 공식으로 해결
−1.01u2−0.14163…u+0.49850…=0
4차 방정식 공식:
위해서 a=−1.01,b=−0.14163…,c=0.49850…u1,2=2(−1.01)−(−0.14163…)±(−0.14163…)2−4(−1.01)⋅0.49850…
u1,2=2(−1.01)−(−0.14163…)±(−0.14163…)2−4(−1.01)⋅0.49850…
(−0.14163…)2−4(−1.01)⋅0.49850…=2.03401…
(−0.14163…)2−4(−1.01)⋅0.49850…
규칙 적용 −(−a)=a=(−0.14163…)2+4⋅1.01⋅0.49850…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.14163…)2=0.14163…2=0.14163…2+4⋅0.49850…⋅1.01
숫자를 곱하시오: 4⋅1.01⋅0.49850…=2.01395…=0.14163…2+2.01395…
0.14163…2=0.02005…=0.02005…+2.01395…
숫자 추가: 0.02005…+2.01395…=2.03401…=2.03401…
u1,2=2(−1.01)−(−0.14163…)±2.03401…
솔루션 분리u1=2(−1.01)−(−0.14163…)+2.03401…,u2=2(−1.01)−(−0.14163…)−2.03401…
u=2(−1.01)−(−0.14163…)+2.03401…:−2.020.14163…+2.03401…
2(−1.01)−(−0.14163…)+2.03401…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.14163…+2.03401…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.14163…+2.03401…
분수 규칙 적용: −ba=−ba=−2.020.14163…+2.03401…
u=2(−1.01)−(−0.14163…)−2.03401…:2.022.03401…−0.14163…
2(−1.01)−(−0.14163…)−2.03401…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.14163…−2.03401…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.14163…−2.03401…
분수 규칙 적용: −b−a=ba0.14163…−2.03401…=−(2.03401…−0.14163…)=2.022.03401…−0.14163…
2차 방정식의 해는 다음과 같다:u=−2.020.14163…+2.03401…,u=2.022.03401…−0.14163…
뒤로 대체 u=cos(A)cos(A)=−2.020.14163…+2.03401…,cos(A)=2.022.03401…−0.14163…
cos(A)=−2.020.14163…+2.03401…,cos(A)=2.022.03401…−0.14163…
cos(A)=−2.020.14163…+2.03401…:A=arccos(−2.020.14163…+2.03401…)+2πn,A=−arccos(−2.020.14163…+2.03401…)+2πn
cos(A)=−2.020.14163…+2.03401…
트리거 역속성 적용
cos(A)=−2.020.14163…+2.03401…
일반 솔루션 cos(A)=−2.020.14163…+2.03401…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnA=arccos(−2.020.14163…+2.03401…)+2πn,A=−arccos(−2.020.14163…+2.03401…)+2πn
A=arccos(−2.020.14163…+2.03401…)+2πn,A=−arccos(−2.020.14163…+2.03401…)+2πn
cos(A)=2.022.03401…−0.14163…:A=arccos(2.022.03401…−0.14163…)+2πn,A=2π−arccos(2.022.03401…−0.14163…)+2πn
cos(A)=2.022.03401…−0.14163…
트리거 역속성 적용
cos(A)=2.022.03401…−0.14163…
일반 솔루션 cos(A)=2.022.03401…−0.14163…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnA=arccos(2.022.03401…−0.14163…)+2πn,A=2π−arccos(2.022.03401…−0.14163…)+2πn
A=arccos(2.022.03401…−0.14163…)+2πn,A=2π−arccos(2.022.03401…−0.14163…)+2πn
모든 솔루션 결합A=arccos(−2.020.14163…+2.03401…)+2πn,A=−arccos(−2.020.14163…+2.03401…)+2πn,A=arccos(2.022.03401…−0.14163…)+2πn,A=2π−arccos(2.022.03401…−0.14163…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(A)−0.1cos(A)=9.86.94
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.020.14163…+2.03401…)+2πn:참
arccos(−2.020.14163…+2.03401…)+2πn
n=1끼우다 arccos(−2.020.14163…+2.03401…)+2π1
sin(A)−0.1cos(A)=9.86.94 위한 {\ quad}끼우다{\ quad} A=arccos(−2.020.14163…+2.03401…)+2π1sin(arccos(−2.020.14163…+2.03401…)+2π1)−0.1cos(arccos(−2.020.14163…+2.03401…)+2π1)=9.86.94
다듬다0.70816…=0.70816…
⇒참
솔루션 확인 −arccos(−2.020.14163…+2.03401…)+2πn:거짓
−arccos(−2.020.14163…+2.03401…)+2πn
n=1끼우다 −arccos(−2.020.14163…+2.03401…)+2π1
sin(A)−0.1cos(A)=9.86.94 위한 {\ quad}끼우다{\ quad} A=−arccos(−2.020.14163…+2.03401…)+2π1sin(−arccos(−2.020.14163…+2.03401…)+2π1)−0.1cos(−arccos(−2.020.14163…+2.03401…)+2π1)=9.86.94
다듬다−0.55293…=0.70816…
⇒거짓
솔루션 확인 arccos(2.022.03401…−0.14163…)+2πn:참
arccos(2.022.03401…−0.14163…)+2πn
n=1끼우다 arccos(2.022.03401…−0.14163…)+2π1
sin(A)−0.1cos(A)=9.86.94 위한 {\ quad}끼우다{\ quad} A=arccos(2.022.03401…−0.14163…)+2π1sin(arccos(2.022.03401…−0.14163…)+2π1)−0.1cos(arccos(2.022.03401…−0.14163…)+2π1)=9.86.94
다듬다0.70816…=0.70816…
⇒참
솔루션 확인 2π−arccos(2.022.03401…−0.14163…)+2πn:거짓
2π−arccos(2.022.03401…−0.14163…)+2πn
n=1끼우다 2π−arccos(2.022.03401…−0.14163…)+2π1
sin(A)−0.1cos(A)=9.86.94 위한 {\ quad}끼우다{\ quad} A=2π−arccos(2.022.03401…−0.14163…)+2π1sin(2π−arccos(2.022.03401…−0.14163…)+2π1)−0.1cos(2π−arccos(2.022.03401…−0.14163…)+2π1)=9.86.94
다듬다−0.83534…=0.70816…
⇒거짓
A=arccos(−2.020.14163…+2.03401…)+2πn,A=arccos(2.022.03401…−0.14163…)+2πn
해를 10진수 형식으로 표시A=2.45933…+2πn,A=0.88159…+2πn