해법
196sin(θ)−49cos(θ)=160
해법
θ=2.47257…+2πn,θ=1.15897…+2πn
+1
도
θ=141.66783…∘+360∘n,θ=66.40464…∘+360∘n솔루션 단계
196sin(θ)−49cos(θ)=160
더하다 49cos(θ) 양쪽으로196sin(θ)=160+49cos(θ)
양쪽을 제곱(196sin(θ))2=(160+49cos(θ))2
빼다 (160+49cos(θ))2 양쪽에서38416sin2(θ)−25600−15680cos(θ)−2401cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−25600−15680cos(θ)−2401cos2(θ)+38416sin2(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ))
−25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ))간소화하다 :−40817cos2(θ)−15680cos(θ)+12816
−25600−15680cos(θ)−2401cos2(θ)+38416(1−cos2(θ))
38416(1−cos2(θ))확대한다:38416−38416cos2(θ)
38416(1−cos2(θ))
분배 법칙 적용: a(b−c)=ab−aca=38416,b=1,c=cos2(θ)=38416⋅1−38416cos2(θ)
숫자를 곱하시오: 38416⋅1=38416=38416−38416cos2(θ)
=−25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ)
−25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ)단순화하세요:−40817cos2(θ)−15680cos(θ)+12816
−25600−15680cos(θ)−2401cos2(θ)+38416−38416cos2(θ)
집단적 용어=−15680cos(θ)−2401cos2(θ)−38416cos2(θ)−25600+38416
유사 요소 추가: −2401cos2(θ)−38416cos2(θ)=−40817cos2(θ)=−15680cos(θ)−40817cos2(θ)−25600+38416
숫자 더하기/ 빼기: −25600+38416=12816=−40817cos2(θ)−15680cos(θ)+12816
=−40817cos2(θ)−15680cos(θ)+12816
=−40817cos2(θ)−15680cos(θ)+12816
12816−15680cos(θ)−40817cos2(θ)=0
대체로 해결
12816−15680cos(θ)−40817cos2(θ)=0
하게: cos(θ)=u12816−15680u−40817u2=0
12816−15680u−40817u2=0:u=−8163415680+2338305088,u=816342338305088−15680
12816−15680u−40817u2=0
표준 양식으로 작성 ax2+bx+c=0−40817u2−15680u+12816=0
쿼드 공식으로 해결
−40817u2−15680u+12816=0
4차 방정식 공식:
위해서 a=−40817,b=−15680,c=12816u1,2=2(−40817)−(−15680)±(−15680)2−4(−40817)⋅12816
u1,2=2(−40817)−(−15680)±(−15680)2−4(−40817)⋅12816
(−15680)2−4(−40817)⋅12816=2338305088
(−15680)2−4(−40817)⋅12816
규칙 적용 −(−a)=a=(−15680)2+4⋅40817⋅12816
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−15680)2=156802=156802+4⋅40817⋅12816
숫자를 곱하시오: 4⋅40817⋅12816=2092442688=156802+2092442688
156802=245862400=245862400+2092442688
숫자 추가: 245862400+2092442688=2338305088=2338305088
u1,2=2(−40817)−(−15680)±2338305088
솔루션 분리u1=2(−40817)−(−15680)+2338305088,u2=2(−40817)−(−15680)−2338305088
u=2(−40817)−(−15680)+2338305088:−8163415680+2338305088
2(−40817)−(−15680)+2338305088
괄호 제거: (−a)=−a,−(−a)=a=−2⋅4081715680+2338305088
숫자를 곱하시오: 2⋅40817=81634=−8163415680+2338305088
분수 규칙 적용: −ba=−ba=−8163415680+2338305088
u=2(−40817)−(−15680)−2338305088:816342338305088−15680
2(−40817)−(−15680)−2338305088
괄호 제거: (−a)=−a,−(−a)=a=−2⋅4081715680−2338305088
숫자를 곱하시오: 2⋅40817=81634=−8163415680−2338305088
분수 규칙 적용: −b−a=ba15680−2338305088=−(2338305088−15680)=816342338305088−15680
2차 방정식의 해는 다음과 같다:u=−8163415680+2338305088,u=816342338305088−15680
뒤로 대체 u=cos(θ)cos(θ)=−8163415680+2338305088,cos(θ)=816342338305088−15680
cos(θ)=−8163415680+2338305088,cos(θ)=816342338305088−15680
cos(θ)=−8163415680+2338305088:θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
cos(θ)=−8163415680+2338305088
트리거 역속성 적용
cos(θ)=−8163415680+2338305088
일반 솔루션 cos(θ)=−8163415680+2338305088cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn
cos(θ)=816342338305088−15680:θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
cos(θ)=816342338305088−15680
트리거 역속성 적용
cos(θ)=816342338305088−15680
일반 솔루션 cos(θ)=816342338305088−15680cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
모든 솔루션 결합θ=arccos(−8163415680+2338305088)+2πn,θ=−arccos(−8163415680+2338305088)+2πn,θ=arccos(816342338305088−15680)+2πn,θ=2π−arccos(816342338305088−15680)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 196sin(θ)−49cos(θ)=160
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−8163415680+2338305088)+2πn:참
arccos(−8163415680+2338305088)+2πn
n=1끼우다 arccos(−8163415680+2338305088)+2π1
196sin(θ)−49cos(θ)=160 위한 {\ quad}끼우다{\ quad} θ=arccos(−8163415680+2338305088)+2π1196sin(arccos(−8163415680+2338305088)+2π1)−49cos(arccos(−8163415680+2338305088)+2π1)=160
다듬다160=160
⇒참
솔루션 확인 −arccos(−8163415680+2338305088)+2πn:거짓
−arccos(−8163415680+2338305088)+2πn
n=1끼우다 −arccos(−8163415680+2338305088)+2π1
196sin(θ)−49cos(θ)=160 위한 {\ quad}끼우다{\ quad} θ=−arccos(−8163415680+2338305088)+2π1196sin(−arccos(−8163415680+2338305088)+2π1)−49cos(−arccos(−8163415680+2338305088)+2π1)=160
다듬다−83.12602…=160
⇒거짓
솔루션 확인 arccos(816342338305088−15680)+2πn:참
arccos(816342338305088−15680)+2πn
n=1끼우다 arccos(816342338305088−15680)+2π1
196sin(θ)−49cos(θ)=160 위한 {\ quad}끼우다{\ quad} θ=arccos(816342338305088−15680)+2π1196sin(arccos(816342338305088−15680)+2π1)−49cos(arccos(816342338305088−15680)+2π1)=160
다듬다160=160
⇒참
솔루션 확인 2π−arccos(816342338305088−15680)+2πn:거짓
2π−arccos(816342338305088−15680)+2πn
n=1끼우다 2π−arccos(816342338305088−15680)+2π1
196sin(θ)−49cos(θ)=160 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(816342338305088−15680)+2π1196sin(2π−arccos(816342338305088−15680)+2π1)−49cos(2π−arccos(816342338305088−15680)+2π1)=160
다듬다−199.22691…=160
⇒거짓
θ=arccos(−8163415680+2338305088)+2πn,θ=arccos(816342338305088−15680)+2πn
해를 10진수 형식으로 표시θ=2.47257…+2πn,θ=1.15897…+2πn