해법
−5cos(x)+8.6602500000000…sin(x)=−5
해법
x=−2.09439…+2πn,x=0+2πn,x=2π−0+2πn
+1
도
x=−119.99997…∘+360∘n,x=0∘+360∘n,x=360∘+360∘n솔루션 단계
−5cos(x)+8.66025…sin(x)=−5
더하다 5cos(x) 양쪽으로8.66025…sin(x)=−5+5cos(x)
양쪽을 제곱(8.66025…sin(x))2=(−5+5cos(x))2
빼다 (−5+5cos(x))2 양쪽에서74.99993…sin2(x)−25+50cos(x)−25cos2(x)=0
삼각성을 사용하여 다시 쓰기
−25−25cos2(x)+50cos(x)+74.99993…sin2(x)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−25−25cos2(x)+50cos(x)+74.99993…(1−cos2(x))
−25−25cos2(x)+50cos(x)+74.99993…(1−cos2(x))간소화하다 :50cos(x)−99.99993…cos2(x)+49.99993…
−25−25cos2(x)+50cos(x)+74.99993…(1−cos2(x))
74.99993…(1−cos2(x))확대한다:74.99993…−74.99993…cos2(x)
74.99993…(1−cos2(x))
분배 법칙 적용: a(b−c)=ab−aca=74.99993…,b=1,c=cos2(x)=74.99993…⋅1−74.99993…cos2(x)
=1⋅74.99993…−74.99993…cos2(x)
숫자를 곱하시오: 1⋅74.99993…=74.99993…=74.99993…−74.99993…cos2(x)
=−25−25cos2(x)+50cos(x)+74.99993…−74.99993…cos2(x)
−25−25cos2(x)+50cos(x)+74.99993…−74.99993…cos2(x)단순화하세요:50cos(x)−99.99993…cos2(x)+49.99993…
−25−25cos2(x)+50cos(x)+74.99993…−74.99993…cos2(x)
집단적 용어=−25cos2(x)+50cos(x)−74.99993…cos2(x)−25+74.99993…
유사 요소 추가: −25cos2(x)−74.99993…cos2(x)=−99.99993…cos2(x)=−99.99993…cos2(x)+50cos(x)−25+74.99993…
숫자 더하기/ 빼기: −25+74.99993…=49.99993…=50cos(x)−99.99993…cos2(x)+49.99993…
=50cos(x)−99.99993…cos2(x)+49.99993…
=50cos(x)−99.99993…cos2(x)+49.99993…
49.99993…+50cos(x)−99.99993…cos2(x)=0
대체로 해결
49.99993…+50cos(x)−99.99993…cos2(x)=0
하게: cos(x)=u49.99993…+50u−99.99993…u2=0
49.99993…+50u−99.99993…u2=0:u=−199.99986…−50+22499.95803…,u=199.99986…50+22499.95803…
49.99993…+50u−99.99993…u2=0
표준 양식으로 작성 ax2+bx+c=0−99.99993…u2+50u+49.99993…=0
쿼드 공식으로 해결
−99.99993…u2+50u+49.99993…=0
4차 방정식 공식:
위해서 a=−99.99993…,b=50,c=49.99993…u1,2=2(−99.99993…)−50±502−4(−99.99993…)⋅49.99993…
u1,2=2(−99.99993…)−50±502−4(−99.99993…)⋅49.99993…
502−4(−99.99993…)⋅49.99993…=22499.95803…
502−4(−99.99993…)⋅49.99993…
규칙 적용 −(−a)=a=502+4⋅99.99993…⋅49.99993…
숫자를 곱하시오: 4⋅99.99993…⋅49.99993…=19999.95803…=502+19999.95803…
502=2500=2500+19999.95803…
숫자 추가: 2500+19999.95803…=22499.95803…=22499.95803…
u1,2=2(−99.99993…)−50±22499.95803…
솔루션 분리u1=2(−99.99993…)−50+22499.95803…,u2=2(−99.99993…)−50−22499.95803…
u=2(−99.99993…)−50+22499.95803…:−199.99986…−50+22499.95803…
2(−99.99993…)−50+22499.95803…
괄호 제거: (−a)=−a=−2⋅99.99993…−50+22499.95803…
숫자를 곱하시오: 2⋅99.99993…=199.99986…=−199.99986…−50+22499.95803…
분수 규칙 적용: −ba=−ba=−199.99986…−50+22499.95803…
u=2(−99.99993…)−50−22499.95803…:199.99986…50+22499.95803…
2(−99.99993…)−50−22499.95803…
괄호 제거: (−a)=−a=−2⋅99.99993…−50−22499.95803…
숫자를 곱하시오: 2⋅99.99993…=199.99986…=−199.99986…−50−22499.95803…
분수 규칙 적용: −b−a=ba−50−22499.95803…=−(50+22499.95803…)=199.99986…50+22499.95803…
2차 방정식의 해는 다음과 같다:u=−199.99986…−50+22499.95803…,u=199.99986…50+22499.95803…
뒤로 대체 u=cos(x)cos(x)=−199.99986…−50+22499.95803…,cos(x)=199.99986…50+22499.95803…
cos(x)=−199.99986…−50+22499.95803…,cos(x)=199.99986…50+22499.95803…
cos(x)=−199.99986…−50+22499.95803…:x=arccos(−199.99986…−50+22499.95803…)+2πn,x=−arccos(−199.99986…−50+22499.95803…)+2πn
cos(x)=−199.99986…−50+22499.95803…
트리거 역속성 적용
cos(x)=−199.99986…−50+22499.95803…
일반 솔루션 cos(x)=−199.99986…−50+22499.95803…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−199.99986…−50+22499.95803…)+2πn,x=−arccos(−199.99986…−50+22499.95803…)+2πn
x=arccos(−199.99986…−50+22499.95803…)+2πn,x=−arccos(−199.99986…−50+22499.95803…)+2πn
cos(x)=199.99986…50+22499.95803…:x=arccos(199.99986…50+22499.95803…)+2πn,x=2π−arccos(199.99986…50+22499.95803…)+2πn
cos(x)=199.99986…50+22499.95803…
트리거 역속성 적용
cos(x)=199.99986…50+22499.95803…
일반 솔루션 cos(x)=199.99986…50+22499.95803…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(199.99986…50+22499.95803…)+2πn,x=2π−arccos(199.99986…50+22499.95803…)+2πn
x=arccos(199.99986…50+22499.95803…)+2πn,x=2π−arccos(199.99986…50+22499.95803…)+2πn
모든 솔루션 결합x=arccos(−199.99986…−50+22499.95803…)+2πn,x=−arccos(−199.99986…−50+22499.95803…)+2πn,x=arccos(199.99986…50+22499.95803…)+2πn,x=2π−arccos(199.99986…50+22499.95803…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 −5cos(x)+8.66025…sin(x)=−5
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−199.99986…−50+22499.95803…)+2πn:거짓
arccos(−199.99986…−50+22499.95803…)+2πn
n=1끼우다 arccos(−199.99986…−50+22499.95803…)+2π1
−5cos(x)+8.66025…sin(x)=−5 위한 {\ quad}끼우다{\ quad} x=arccos(−199.99986…−50+22499.95803…)+2π1−5cos(arccos(−199.99986…−50+22499.95803…)+2π1)+8.66025…sin(arccos(−199.99986…−50+22499.95803…)+2π1)=−5
다듬다9.99999…=−5
⇒거짓
솔루션 확인 −arccos(−199.99986…−50+22499.95803…)+2πn:참
−arccos(−199.99986…−50+22499.95803…)+2πn
n=1끼우다 −arccos(−199.99986…−50+22499.95803…)+2π1
−5cos(x)+8.66025…sin(x)=−5 위한 {\ quad}끼우다{\ quad} x=−arccos(−199.99986…−50+22499.95803…)+2π1−5cos(−arccos(−199.99986…−50+22499.95803…)+2π1)+8.66025…sin(−arccos(−199.99986…−50+22499.95803…)+2π1)=−5
다듬다−5=−5
⇒참
솔루션 확인 arccos(199.99986…50+22499.95803…)+2πn:참
arccos(199.99986…50+22499.95803…)+2πn
n=1끼우다 arccos(199.99986…50+22499.95803…)+2π1
−5cos(x)+8.66025…sin(x)=−5 위한 {\ quad}끼우다{\ quad} x=arccos(199.99986…50+22499.95803…)+2π1−5cos(arccos(199.99986…50+22499.95803…)+2π1)+8.66025…sin(arccos(199.99986…50+22499.95803…)+2π1)=−5
다듬다−5=−5
⇒참
솔루션 확인 2π−arccos(199.99986…50+22499.95803…)+2πn:참
2π−arccos(199.99986…50+22499.95803…)+2πn
n=1끼우다 2π−arccos(199.99986…50+22499.95803…)+2π1
−5cos(x)+8.66025…sin(x)=−5 위한 {\ quad}끼우다{\ quad} x=2π−arccos(199.99986…50+22499.95803…)+2π1−5cos(2π−arccos(199.99986…50+22499.95803…)+2π1)+8.66025…sin(2π−arccos(199.99986…50+22499.95803…)+2π1)=−5
다듬다−5=−5
⇒참
x=−arccos(−199.99986…−50+22499.95803…)+2πn,x=arccos(199.99986…50+22499.95803…)+2πn,x=2π−arccos(199.99986…50+22499.95803…)+2πn
해를 10진수 형식으로 표시x=−2.09439…+2πn,x=0+2πn,x=2π−0+2πn