해법
117.72sin(θ)−35.316cos(θ)−12.5=0
해법
θ=0.39333…+2πn,θ=π+0.18957…+2πn
+1
도
θ=22.53666…∘+360∘n,θ=190.86182…∘+360∘n솔루션 단계
117.72sin(θ)−35.316cos(θ)−12.5=0
더하다 35.316cos(θ) 양쪽으로117.72sin(θ)−12.5=35.316cos(θ)
양쪽을 제곱(117.72sin(θ)−12.5)2=(35.316cos(θ))2
빼다 (35.316cos(θ))2 양쪽에서(117.72sin(θ)−12.5)2−1247.219856cos2(θ)=0
삼각성을 사용하여 다시 쓰기
(−12.5+117.72sin(θ))2−1247.219856cos2(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(−12.5+117.72sin(θ))2−1247.219856(1−sin2(θ))
(−12.5+117.72sin(θ))2−1247.219856(1−sin2(θ))간소화하다 :15105.218256sin2(θ)−2943sin(θ)−1090.969856
(−12.5+117.72sin(θ))2−1247.219856(1−sin2(θ))
(−12.5+117.72sin(θ))2:156.25−2943sin(θ)+13857.9984sin2(θ)
완벽한 정사각형 공식 적용: (a+b)2=a2+2ab+b2a=−12.5,b=117.72sin(θ)
=(−12.5)2+2(−12.5)⋅117.72sin(θ)+(117.72sin(θ))2
(−12.5)2+2(−12.5)⋅117.72sin(θ)+(117.72sin(θ))2단순화하세요:156.25−2943sin(θ)+13857.9984sin2(θ)
(−12.5)2+2(−12.5)⋅117.72sin(θ)+(117.72sin(θ))2
괄호 제거: (−a)=−a=(−12.5)2−2⋅12.5⋅117.72sin(θ)+(117.72sin(θ))2
(−12.5)2=156.25
(−12.5)2
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−12.5)2=12.52=12.52
12.52=156.25=156.25
2⋅12.5⋅117.72sin(θ)=2943sin(θ)
2⋅12.5⋅117.72sin(θ)
숫자를 곱하시오: 2⋅12.5⋅117.72=2943=2943sin(θ)
(117.72sin(θ))2=13857.9984sin2(θ)
(117.72sin(θ))2
지수 규칙 적용: (a⋅b)n=anbn=117.722sin2(θ)
117.722=13857.9984=13857.9984sin2(θ)
=156.25−2943sin(θ)+13857.9984sin2(θ)
=156.25−2943sin(θ)+13857.9984sin2(θ)
=156.25−2943sin(θ)+13857.9984sin2(θ)−1247.219856(1−sin2(θ))
−1247.219856(1−sin2(θ))확대한다:−1247.219856+1247.219856sin2(θ)
−1247.219856(1−sin2(θ))
분배 법칙 적용: a(b−c)=ab−aca=−1247.219856,b=1,c=sin2(θ)=−1247.219856⋅1−(−1247.219856)sin2(θ)
마이너스 플러스 규칙 적용−(−a)=a=−1⋅1247.219856+1247.219856sin2(θ)
숫자를 곱하시오: 1⋅1247.219856=1247.219856=−1247.219856+1247.219856sin2(θ)
=156.25−2943sin(θ)+13857.9984sin2(θ)−1247.219856+1247.219856sin2(θ)
156.25−2943sin(θ)+13857.9984sin2(θ)−1247.219856+1247.219856sin2(θ)단순화하세요:15105.218256sin2(θ)−2943sin(θ)−1090.969856
156.25−2943sin(θ)+13857.9984sin2(θ)−1247.219856+1247.219856sin2(θ)
집단적 용어=−2943sin(θ)+13857.9984sin2(θ)+1247.219856sin2(θ)+156.25−1247.219856
유사 요소 추가: 13857.9984sin2(θ)+1247.219856sin2(θ)=15105.218256sin2(θ)=−2943sin(θ)+15105.218256sin2(θ)+156.25−1247.219856
숫자 더하기/ 빼기: 156.25−1247.219856=−1090.969856=15105.218256sin2(θ)−2943sin(θ)−1090.969856
=15105.218256sin2(θ)−2943sin(θ)−1090.969856
=15105.218256sin2(θ)−2943sin(θ)−1090.969856
−1090.969856+15105.218256sin2(θ)−2943sin(θ)=0
대체로 해결
−1090.969856+15105.218256sin2(θ)−2943sin(θ)=0
하게: sin(θ)=u−1090.969856+15105.218256u2−2943u=0
−1090.969856+15105.218256u2−2943u=0:u=20.19483…+0.32685…,u=20.19483…−0.32685…
−1090.969856+15105.218256u2−2943u=0
양쪽을 다음으로 나눕니다 15105.218256−15105.2182561090.969856+15105.21825615105.218256u2−15105.2182562943u=15105.2182560
표준 양식으로 작성 ax2+bx+c=0u2−0.19483…u−0.07222…=0
쿼드 공식으로 해결
u2−0.19483…u−0.07222…=0
4차 방정식 공식:
위해서 a=1,b=−0.19483…,c=−0.07222…u1,2=2⋅1−(−0.19483…)±(−0.19483…)2−4⋅1⋅(−0.07222…)
u1,2=2⋅1−(−0.19483…)±(−0.19483…)2−4⋅1⋅(−0.07222…)
(−0.19483…)2−4⋅1⋅(−0.07222…)=0.32685…
(−0.19483…)2−4⋅1⋅(−0.07222…)
규칙 적용 −(−a)=a=(−0.19483…)2+4⋅1⋅0.07222…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.19483…)2=0.19483…2=0.19483…2+4⋅1⋅0.07222…
숫자를 곱하시오: 4⋅1⋅0.07222…=0.28889…=0.19483…2+0.28889…
0.19483…2=0.03796…=0.03796…+0.28889…
숫자 추가: 0.03796…+0.28889…=0.32685…=0.32685…
u1,2=2⋅1−(−0.19483…)±0.32685…
솔루션 분리u1=2⋅1−(−0.19483…)+0.32685…,u2=2⋅1−(−0.19483…)−0.32685…
u=2⋅1−(−0.19483…)+0.32685…:20.19483…+0.32685…
2⋅1−(−0.19483…)+0.32685…
규칙 적용 −(−a)=a=2⋅10.19483…+0.32685…
숫자를 곱하시오: 2⋅1=2=20.19483…+0.32685…
u=2⋅1−(−0.19483…)−0.32685…:20.19483…−0.32685…
2⋅1−(−0.19483…)−0.32685…
규칙 적용 −(−a)=a=2⋅10.19483…−0.32685…
숫자를 곱하시오: 2⋅1=2=20.19483…−0.32685…
2차 방정식의 해는 다음과 같다:u=20.19483…+0.32685…,u=20.19483…−0.32685…
뒤로 대체 u=sin(θ)sin(θ)=20.19483…+0.32685…,sin(θ)=20.19483…−0.32685…
sin(θ)=20.19483…+0.32685…,sin(θ)=20.19483…−0.32685…
sin(θ)=20.19483…+0.32685…:θ=arcsin(20.19483…+0.32685…)+2πn,θ=π−arcsin(20.19483…+0.32685…)+2πn
sin(θ)=20.19483…+0.32685…
트리거 역속성 적용
sin(θ)=20.19483…+0.32685…
일반 솔루션 sin(θ)=20.19483…+0.32685…sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(20.19483…+0.32685…)+2πn,θ=π−arcsin(20.19483…+0.32685…)+2πn
θ=arcsin(20.19483…+0.32685…)+2πn,θ=π−arcsin(20.19483…+0.32685…)+2πn
sin(θ)=20.19483…−0.32685…:θ=arcsin(20.19483…−0.32685…)+2πn,θ=π+arcsin(−20.19483…−0.32685…)+2πn
sin(θ)=20.19483…−0.32685…
트리거 역속성 적용
sin(θ)=20.19483…−0.32685…
일반 솔루션 sin(θ)=20.19483…−0.32685…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnθ=arcsin(20.19483…−0.32685…)+2πn,θ=π+arcsin(−20.19483…−0.32685…)+2πn
θ=arcsin(20.19483…−0.32685…)+2πn,θ=π+arcsin(−20.19483…−0.32685…)+2πn
모든 솔루션 결합θ=arcsin(20.19483…+0.32685…)+2πn,θ=π−arcsin(20.19483…+0.32685…)+2πn,θ=arcsin(20.19483…−0.32685…)+2πn,θ=π+arcsin(−20.19483…−0.32685…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 117.72sin(θ)−35.316cos(θ)−12.5=0
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arcsin(20.19483…+0.32685…)+2πn:참
arcsin(20.19483…+0.32685…)+2πn
n=1끼우다 arcsin(20.19483…+0.32685…)+2π1
117.72sin(θ)−35.316cos(θ)−12.5=0 위한 {\ quad}끼우다{\ quad} θ=arcsin(20.19483…+0.32685…)+2π1117.72sin(arcsin(20.19483…+0.32685…)+2π1)−35.316cos(arcsin(20.19483…+0.32685…)+2π1)−12.5=0
다듬다0=0
⇒참
솔루션 확인 π−arcsin(20.19483…+0.32685…)+2πn:거짓
π−arcsin(20.19483…+0.32685…)+2πn
n=1끼우다 π−arcsin(20.19483…+0.32685…)+2π1
117.72sin(θ)−35.316cos(θ)−12.5=0 위한 {\ quad}끼우다{\ quad} θ=π−arcsin(20.19483…+0.32685…)+2π1117.72sin(π−arcsin(20.19483…+0.32685…)+2π1)−35.316cos(π−arcsin(20.19483…+0.32685…)+2π1)−12.5=0
다듬다65.23815…=0
⇒거짓
솔루션 확인 arcsin(20.19483…−0.32685…)+2πn:거짓
arcsin(20.19483…−0.32685…)+2πn
n=1끼우다 arcsin(20.19483…−0.32685…)+2π1
117.72sin(θ)−35.316cos(θ)−12.5=0 위한 {\ quad}끼우다{\ quad} θ=arcsin(20.19483…−0.32685…)+2π1117.72sin(arcsin(20.19483…−0.32685…)+2π1)−35.316cos(arcsin(20.19483…−0.32685…)+2π1)−12.5=0
다듬다−69.36659…=0
⇒거짓
솔루션 확인 π+arcsin(−20.19483…−0.32685…)+2πn:참
π+arcsin(−20.19483…−0.32685…)+2πn
n=1끼우다 π+arcsin(−20.19483…−0.32685…)+2π1
117.72sin(θ)−35.316cos(θ)−12.5=0 위한 {\ quad}끼우다{\ quad} θ=π+arcsin(−20.19483…−0.32685…)+2π1117.72sin(π+arcsin(−20.19483…−0.32685…)+2π1)−35.316cos(π+arcsin(−20.19483…−0.32685…)+2π1)−12.5=0
다듬다0=0
⇒참
θ=arcsin(20.19483…+0.32685…)+2πn,θ=π+arcsin(−20.19483…−0.32685…)+2πn
해를 10진수 형식으로 표시θ=0.39333…+2πn,θ=π+0.18957…+2πn