해법
sin(θ)−0.2cos(θ)=9.86.25
해법
θ=2.66338…+2πn,θ=0.87300…+2πn
+1
도
θ=152.60055…∘+360∘n,θ=50.01930…∘+360∘n솔루션 단계
sin(θ)−0.2cos(θ)=9.86.25
더하다 0.2cos(θ) 양쪽으로sin(θ)=0.63775…+0.2cos(θ)
양쪽을 제곱sin2(θ)=(0.63775…+0.2cos(θ))2
빼다 (0.63775…+0.2cos(θ))2 양쪽에서sin2(θ)−0.40673…−0.25510…cos(θ)−0.04cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−0.40673…+sin2(θ)−0.04cos2(θ)−0.25510…cos(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ)
−0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ)간소화하다 :−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
−0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ)
유사 요소 추가: −cos2(θ)−0.04cos2(θ)=−1.04cos2(θ)=−0.40673…+1−1.04cos2(θ)−0.25510…cos(θ)
숫자 더하기/ 빼기: −0.40673…+1=0.59326…=−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
=−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
0.59326…−0.25510…cos(θ)−1.04cos2(θ)=0
대체로 해결
0.59326…−0.25510…cos(θ)−1.04cos2(θ)=0
하게: cos(θ)=u0.59326…−0.25510…u−1.04u2=0
0.59326…−0.25510…u−1.04u2=0:u=−2.080.25510…+2.53307…,u=2.082.53307…−0.25510…
0.59326…−0.25510…u−1.04u2=0
표준 양식으로 작성 ax2+bx+c=0−1.04u2−0.25510…u+0.59326…=0
쿼드 공식으로 해결
−1.04u2−0.25510…u+0.59326…=0
4차 방정식 공식:
위해서 a=−1.04,b=−0.25510…,c=0.59326…u1,2=2(−1.04)−(−0.25510…)±(−0.25510…)2−4(−1.04)⋅0.59326…
u1,2=2(−1.04)−(−0.25510…)±(−0.25510…)2−4(−1.04)⋅0.59326…
(−0.25510…)2−4(−1.04)⋅0.59326…=2.53307…
(−0.25510…)2−4(−1.04)⋅0.59326…
규칙 적용 −(−a)=a=(−0.25510…)2+4⋅1.04⋅0.59326…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.25510…)2=0.25510…2=0.25510…2+4⋅0.59326…⋅1.04
숫자를 곱하시오: 4⋅1.04⋅0.59326…=2.46799…=0.25510…2+2.46799…
0.25510…2=0.06507…=0.06507…+2.46799…
숫자 추가: 0.06507…+2.46799…=2.53307…=2.53307…
u1,2=2(−1.04)−(−0.25510…)±2.53307…
솔루션 분리u1=2(−1.04)−(−0.25510…)+2.53307…,u2=2(−1.04)−(−0.25510…)−2.53307…
u=2(−1.04)−(−0.25510…)+2.53307…:−2.080.25510…+2.53307…
2(−1.04)−(−0.25510…)+2.53307…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.25510…+2.53307…
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.25510…+2.53307…
분수 규칙 적용: −ba=−ba=−2.080.25510…+2.53307…
u=2(−1.04)−(−0.25510…)−2.53307…:2.082.53307…−0.25510…
2(−1.04)−(−0.25510…)−2.53307…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.25510…−2.53307…
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.25510…−2.53307…
분수 규칙 적용: −b−a=ba0.25510…−2.53307…=−(2.53307…−0.25510…)=2.082.53307…−0.25510…
2차 방정식의 해는 다음과 같다:u=−2.080.25510…+2.53307…,u=2.082.53307…−0.25510…
뒤로 대체 u=cos(θ)cos(θ)=−2.080.25510…+2.53307…,cos(θ)=2.082.53307…−0.25510…
cos(θ)=−2.080.25510…+2.53307…,cos(θ)=2.082.53307…−0.25510…
cos(θ)=−2.080.25510…+2.53307…:θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
cos(θ)=−2.080.25510…+2.53307…
트리거 역속성 적용
cos(θ)=−2.080.25510…+2.53307…
일반 솔루션 cos(θ)=−2.080.25510…+2.53307…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
cos(θ)=2.082.53307…−0.25510…:θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
cos(θ)=2.082.53307…−0.25510…
트리거 역속성 적용
cos(θ)=2.082.53307…−0.25510…
일반 솔루션 cos(θ)=2.082.53307…−0.25510…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
모든 솔루션 결합θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn,θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(θ)−0.2cos(θ)=9.86.25
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.080.25510…+2.53307…)+2πn:참
arccos(−2.080.25510…+2.53307…)+2πn
n=1끼우다 arccos(−2.080.25510…+2.53307…)+2π1
sin(θ)−0.2cos(θ)=9.86.25 위한 {\ quad}끼우다{\ quad} θ=arccos(−2.080.25510…+2.53307…)+2π1sin(arccos(−2.080.25510…+2.53307…)+2π1)−0.2cos(arccos(−2.080.25510…+2.53307…)+2π1)=9.86.25
다듬다0.63775…=0.63775…
⇒참
솔루션 확인 −arccos(−2.080.25510…+2.53307…)+2πn:거짓
−arccos(−2.080.25510…+2.53307…)+2πn
n=1끼우다 −arccos(−2.080.25510…+2.53307…)+2π1
sin(θ)−0.2cos(θ)=9.86.25 위한 {\ quad}끼우다{\ quad} θ=−arccos(−2.080.25510…+2.53307…)+2π1sin(−arccos(−2.080.25510…+2.53307…)+2π1)−0.2cos(−arccos(−2.080.25510…+2.53307…)+2π1)=9.86.25
다듬다−0.28262…=0.63775…
⇒거짓
솔루션 확인 arccos(2.082.53307…−0.25510…)+2πn:참
arccos(2.082.53307…−0.25510…)+2πn
n=1끼우다 arccos(2.082.53307…−0.25510…)+2π1
sin(θ)−0.2cos(θ)=9.86.25 위한 {\ quad}끼우다{\ quad} θ=arccos(2.082.53307…−0.25510…)+2π1sin(arccos(2.082.53307…−0.25510…)+2π1)−0.2cos(arccos(2.082.53307…−0.25510…)+2π1)=9.86.25
다듬다0.63775…=0.63775…
⇒참
솔루션 확인 2π−arccos(2.082.53307…−0.25510…)+2πn:거짓
2π−arccos(2.082.53307…−0.25510…)+2πn
n=1끼우다 2π−arccos(2.082.53307…−0.25510…)+2π1
sin(θ)−0.2cos(θ)=9.86.25 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(2.082.53307…−0.25510…)+2π1sin(2π−arccos(2.082.53307…−0.25510…)+2π1)−0.2cos(2π−arccos(2.082.53307…−0.25510…)+2π1)=9.86.25
다듬다−0.89476…=0.63775…
⇒거짓
θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=arccos(2.082.53307…−0.25510…)+2πn
해를 10진수 형식으로 표시θ=2.66338…+2πn,θ=0.87300…+2πn