해법
sin(θ)−0.1cos(θ)=9.86.88
해법
θ=2.46788…+2πn,θ=0.87304…+2πn
+1
도
θ=141.39926…∘+360∘n,θ=50.02191…∘+360∘n솔루션 단계
sin(θ)−0.1cos(θ)=9.86.88
더하다 0.1cos(θ) 양쪽으로sin(θ)=0.70204…+0.1cos(θ)
양쪽을 제곱sin2(θ)=(0.70204…+0.1cos(θ))2
빼다 (0.70204…+0.1cos(θ))2 양쪽에서sin2(θ)−0.49286…−0.14040…cos(θ)−0.01cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−0.49286…+sin2(θ)−0.01cos2(θ)−0.14040…cos(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.49286…+1−cos2(θ)−0.01cos2(θ)−0.14040…cos(θ)
−0.49286…+1−cos2(θ)−0.01cos2(θ)−0.14040…cos(θ)간소화하다 :−1.01cos2(θ)−0.14040…cos(θ)+0.50713…
−0.49286…+1−cos2(θ)−0.01cos2(θ)−0.14040…cos(θ)
유사 요소 추가: −cos2(θ)−0.01cos2(θ)=−1.01cos2(θ)=−0.49286…+1−1.01cos2(θ)−0.14040…cos(θ)
숫자 더하기/ 빼기: −0.49286…+1=0.50713…=−1.01cos2(θ)−0.14040…cos(θ)+0.50713…
=−1.01cos2(θ)−0.14040…cos(θ)+0.50713…
0.50713…−0.14040…cos(θ)−1.01cos2(θ)=0
대체로 해결
0.50713…−0.14040…cos(θ)−1.01cos2(θ)=0
하게: cos(θ)=u0.50713…−0.14040…u−1.01u2=0
0.50713…−0.14040…u−1.01u2=0:u=−2.020.14040…+2.06855…,u=2.022.06855…−0.14040…
0.50713…−0.14040…u−1.01u2=0
표준 양식으로 작성 ax2+bx+c=0−1.01u2−0.14040…u+0.50713…=0
쿼드 공식으로 해결
−1.01u2−0.14040…u+0.50713…=0
4차 방정식 공식:
위해서 a=−1.01,b=−0.14040…,c=0.50713…u1,2=2(−1.01)−(−0.14040…)±(−0.14040…)2−4(−1.01)⋅0.50713…
u1,2=2(−1.01)−(−0.14040…)±(−0.14040…)2−4(−1.01)⋅0.50713…
(−0.14040…)2−4(−1.01)⋅0.50713…=2.06855…
(−0.14040…)2−4(−1.01)⋅0.50713…
규칙 적용 −(−a)=a=(−0.14040…)2+4⋅1.01⋅0.50713…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.14040…)2=0.14040…2=0.14040…2+4⋅0.50713…⋅1.01
숫자를 곱하시오: 4⋅1.01⋅0.50713…=2.04884…=0.14040…2+2.04884…
0.14040…2=0.01971…=0.01971…+2.04884…
숫자 추가: 0.01971…+2.04884…=2.06855…=2.06855…
u1,2=2(−1.01)−(−0.14040…)±2.06855…
솔루션 분리u1=2(−1.01)−(−0.14040…)+2.06855…,u2=2(−1.01)−(−0.14040…)−2.06855…
u=2(−1.01)−(−0.14040…)+2.06855…:−2.020.14040…+2.06855…
2(−1.01)−(−0.14040…)+2.06855…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.14040…+2.06855…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.14040…+2.06855…
분수 규칙 적용: −ba=−ba=−2.020.14040…+2.06855…
u=2(−1.01)−(−0.14040…)−2.06855…:2.022.06855…−0.14040…
2(−1.01)−(−0.14040…)−2.06855…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.010.14040…−2.06855…
숫자를 곱하시오: 2⋅1.01=2.02=−2.020.14040…−2.06855…
분수 규칙 적용: −b−a=ba0.14040…−2.06855…=−(2.06855…−0.14040…)=2.022.06855…−0.14040…
2차 방정식의 해는 다음과 같다:u=−2.020.14040…+2.06855…,u=2.022.06855…−0.14040…
뒤로 대체 u=cos(θ)cos(θ)=−2.020.14040…+2.06855…,cos(θ)=2.022.06855…−0.14040…
cos(θ)=−2.020.14040…+2.06855…,cos(θ)=2.022.06855…−0.14040…
cos(θ)=−2.020.14040…+2.06855…:θ=arccos(−2.020.14040…+2.06855…)+2πn,θ=−arccos(−2.020.14040…+2.06855…)+2πn
cos(θ)=−2.020.14040…+2.06855…
트리거 역속성 적용
cos(θ)=−2.020.14040…+2.06855…
일반 솔루션 cos(θ)=−2.020.14040…+2.06855…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.020.14040…+2.06855…)+2πn,θ=−arccos(−2.020.14040…+2.06855…)+2πn
θ=arccos(−2.020.14040…+2.06855…)+2πn,θ=−arccos(−2.020.14040…+2.06855…)+2πn
cos(θ)=2.022.06855…−0.14040…:θ=arccos(2.022.06855…−0.14040…)+2πn,θ=2π−arccos(2.022.06855…−0.14040…)+2πn
cos(θ)=2.022.06855…−0.14040…
트리거 역속성 적용
cos(θ)=2.022.06855…−0.14040…
일반 솔루션 cos(θ)=2.022.06855…−0.14040…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.022.06855…−0.14040…)+2πn,θ=2π−arccos(2.022.06855…−0.14040…)+2πn
θ=arccos(2.022.06855…−0.14040…)+2πn,θ=2π−arccos(2.022.06855…−0.14040…)+2πn
모든 솔루션 결합θ=arccos(−2.020.14040…+2.06855…)+2πn,θ=−arccos(−2.020.14040…+2.06855…)+2πn,θ=arccos(2.022.06855…−0.14040…)+2πn,θ=2π−arccos(2.022.06855…−0.14040…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(θ)−0.1cos(θ)=9.86.88
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.020.14040…+2.06855…)+2πn:참
arccos(−2.020.14040…+2.06855…)+2πn
n=1끼우다 arccos(−2.020.14040…+2.06855…)+2π1
sin(θ)−0.1cos(θ)=9.86.88 위한 {\ quad}끼우다{\ quad} θ=arccos(−2.020.14040…+2.06855…)+2π1sin(arccos(−2.020.14040…+2.06855…)+2π1)−0.1cos(arccos(−2.020.14040…+2.06855…)+2π1)=9.86.88
다듬다0.70204…=0.70204…
⇒참
솔루션 확인 −arccos(−2.020.14040…+2.06855…)+2πn:거짓
−arccos(−2.020.14040…+2.06855…)+2πn
n=1끼우다 −arccos(−2.020.14040…+2.06855…)+2π1
sin(θ)−0.1cos(θ)=9.86.88 위한 {\ quad}끼우다{\ quad} θ=−arccos(−2.020.14040…+2.06855…)+2π1sin(−arccos(−2.020.14040…+2.06855…)+2π1)−0.1cos(−arccos(−2.020.14040…+2.06855…)+2π1)=9.86.88
다듬다−0.54573…=0.70204…
⇒거짓
솔루션 확인 arccos(2.022.06855…−0.14040…)+2πn:참
arccos(2.022.06855…−0.14040…)+2πn
n=1끼우다 arccos(2.022.06855…−0.14040…)+2π1
sin(θ)−0.1cos(θ)=9.86.88 위한 {\ quad}끼우다{\ quad} θ=arccos(2.022.06855…−0.14040…)+2π1sin(arccos(2.022.06855…−0.14040…)+2π1)−0.1cos(arccos(2.022.06855…−0.14040…)+2π1)=9.86.88
다듬다0.70204…=0.70204…
⇒참
솔루션 확인 2π−arccos(2.022.06855…−0.14040…)+2πn:거짓
2π−arccos(2.022.06855…−0.14040…)+2πn
n=1끼우다 2π−arccos(2.022.06855…−0.14040…)+2π1
sin(θ)−0.1cos(θ)=9.86.88 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(2.022.06855…−0.14040…)+2π1sin(2π−arccos(2.022.06855…−0.14040…)+2π1)−0.1cos(2π−arccos(2.022.06855…−0.14040…)+2π1)=9.86.88
다듬다−0.83053…=0.70204…
⇒거짓
θ=arccos(−2.020.14040…+2.06855…)+2πn,θ=arccos(2.022.06855…−0.14040…)+2πn
해를 10진수 형식으로 표시θ=2.46788…+2πn,θ=0.87304…+2πn