해법
sin(θ)−0.2cos(θ)=9.87.51
해법
θ=2.48873…+2πn,θ=1.04764…+2πn
+1
도
θ=142.59422…∘+360∘n,θ=60.02563…∘+360∘n솔루션 단계
sin(θ)−0.2cos(θ)=9.87.51
더하다 0.2cos(θ) 양쪽으로sin(θ)=0.76632…+0.2cos(θ)
양쪽을 제곱sin2(θ)=(0.76632…+0.2cos(θ))2
빼다 (0.76632…+0.2cos(θ))2 양쪽에서sin2(θ)−0.58725…−0.30653…cos(θ)−0.04cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−0.58725…+sin2(θ)−0.04cos2(θ)−0.30653…cos(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.58725…+1−cos2(θ)−0.04cos2(θ)−0.30653…cos(θ)
−0.58725…+1−cos2(θ)−0.04cos2(θ)−0.30653…cos(θ)간소화하다 :−1.04cos2(θ)−0.30653…cos(θ)+0.41274…
−0.58725…+1−cos2(θ)−0.04cos2(θ)−0.30653…cos(θ)
유사 요소 추가: −cos2(θ)−0.04cos2(θ)=−1.04cos2(θ)=−0.58725…+1−1.04cos2(θ)−0.30653…cos(θ)
숫자 더하기/ 빼기: −0.58725…+1=0.41274…=−1.04cos2(θ)−0.30653…cos(θ)+0.41274…
=−1.04cos2(θ)−0.30653…cos(θ)+0.41274…
0.41274…−0.30653…cos(θ)−1.04cos2(θ)=0
대체로 해결
0.41274…−0.30653…cos(θ)−1.04cos2(θ)=0
하게: cos(θ)=u0.41274…−0.30653…u−1.04u2=0
0.41274…−0.30653…u−1.04u2=0:u=−2.080.30653…+1.81097…,u=2.081.81097…−0.30653…
0.41274…−0.30653…u−1.04u2=0
표준 양식으로 작성 ax2+bx+c=0−1.04u2−0.30653…u+0.41274…=0
쿼드 공식으로 해결
−1.04u2−0.30653…u+0.41274…=0
4차 방정식 공식:
위해서 a=−1.04,b=−0.30653…,c=0.41274…u1,2=2(−1.04)−(−0.30653…)±(−0.30653…)2−4(−1.04)⋅0.41274…
u1,2=2(−1.04)−(−0.30653…)±(−0.30653…)2−4(−1.04)⋅0.41274…
(−0.30653…)2−4(−1.04)⋅0.41274…=1.81097…
(−0.30653…)2−4(−1.04)⋅0.41274…
규칙 적용 −(−a)=a=(−0.30653…)2+4⋅1.04⋅0.41274…
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.30653…)2=0.30653…2=0.30653…2+4⋅0.41274…⋅1.04
숫자를 곱하시오: 4⋅1.04⋅0.41274…=1.71701…=0.30653…2+1.71701…
0.30653…2=0.09396…=0.09396…+1.71701…
숫자 추가: 0.09396…+1.71701…=1.81097…=1.81097…
u1,2=2(−1.04)−(−0.30653…)±1.81097…
솔루션 분리u1=2(−1.04)−(−0.30653…)+1.81097…,u2=2(−1.04)−(−0.30653…)−1.81097…
u=2(−1.04)−(−0.30653…)+1.81097…:−2.080.30653…+1.81097…
2(−1.04)−(−0.30653…)+1.81097…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.30653…+1.81097…
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.30653…+1.81097…
분수 규칙 적용: −ba=−ba=−2.080.30653…+1.81097…
u=2(−1.04)−(−0.30653…)−1.81097…:2.081.81097…−0.30653…
2(−1.04)−(−0.30653…)−1.81097…
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.30653…−1.81097…
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.30653…−1.81097…
분수 규칙 적용: −b−a=ba0.30653…−1.81097…=−(1.81097…−0.30653…)=2.081.81097…−0.30653…
2차 방정식의 해는 다음과 같다:u=−2.080.30653…+1.81097…,u=2.081.81097…−0.30653…
뒤로 대체 u=cos(θ)cos(θ)=−2.080.30653…+1.81097…,cos(θ)=2.081.81097…−0.30653…
cos(θ)=−2.080.30653…+1.81097…,cos(θ)=2.081.81097…−0.30653…
cos(θ)=−2.080.30653…+1.81097…:θ=arccos(−2.080.30653…+1.81097…)+2πn,θ=−arccos(−2.080.30653…+1.81097…)+2πn
cos(θ)=−2.080.30653…+1.81097…
트리거 역속성 적용
cos(θ)=−2.080.30653…+1.81097…
일반 솔루션 cos(θ)=−2.080.30653…+1.81097…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.080.30653…+1.81097…)+2πn,θ=−arccos(−2.080.30653…+1.81097…)+2πn
θ=arccos(−2.080.30653…+1.81097…)+2πn,θ=−arccos(−2.080.30653…+1.81097…)+2πn
cos(θ)=2.081.81097…−0.30653…:θ=arccos(2.081.81097…−0.30653…)+2πn,θ=2π−arccos(2.081.81097…−0.30653…)+2πn
cos(θ)=2.081.81097…−0.30653…
트리거 역속성 적용
cos(θ)=2.081.81097…−0.30653…
일반 솔루션 cos(θ)=2.081.81097…−0.30653…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.081.81097…−0.30653…)+2πn,θ=2π−arccos(2.081.81097…−0.30653…)+2πn
θ=arccos(2.081.81097…−0.30653…)+2πn,θ=2π−arccos(2.081.81097…−0.30653…)+2πn
모든 솔루션 결합θ=arccos(−2.080.30653…+1.81097…)+2πn,θ=−arccos(−2.080.30653…+1.81097…)+2πn,θ=arccos(2.081.81097…−0.30653…)+2πn,θ=2π−arccos(2.081.81097…−0.30653…)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(θ)−0.2cos(θ)=9.87.51
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.080.30653…+1.81097…)+2πn:참
arccos(−2.080.30653…+1.81097…)+2πn
n=1끼우다 arccos(−2.080.30653…+1.81097…)+2π1
sin(θ)−0.2cos(θ)=9.87.51 위한 {\ quad}끼우다{\ quad} θ=arccos(−2.080.30653…+1.81097…)+2π1sin(arccos(−2.080.30653…+1.81097…)+2π1)−0.2cos(arccos(−2.080.30653…+1.81097…)+2π1)=9.87.51
다듬다0.76632…=0.76632…
⇒참
솔루션 확인 −arccos(−2.080.30653…+1.81097…)+2πn:거짓
−arccos(−2.080.30653…+1.81097…)+2πn
n=1끼우다 −arccos(−2.080.30653…+1.81097…)+2π1
sin(θ)−0.2cos(θ)=9.87.51 위한 {\ quad}끼우다{\ quad} θ=−arccos(−2.080.30653…+1.81097…)+2π1sin(−arccos(−2.080.30653…+1.81097…)+2π1)−0.2cos(−arccos(−2.080.30653…+1.81097…)+2π1)=9.87.51
다듬다−0.44858…=0.76632…
⇒거짓
솔루션 확인 arccos(2.081.81097…−0.30653…)+2πn:참
arccos(2.081.81097…−0.30653…)+2πn
n=1끼우다 arccos(2.081.81097…−0.30653…)+2π1
sin(θ)−0.2cos(θ)=9.87.51 위한 {\ quad}끼우다{\ quad} θ=arccos(2.081.81097…−0.30653…)+2π1sin(arccos(2.081.81097…−0.30653…)+2π1)−0.2cos(arccos(2.081.81097…−0.30653…)+2π1)=9.87.51
다듬다0.76632…=0.76632…
⇒참
솔루션 확인 2π−arccos(2.081.81097…−0.30653…)+2πn:거짓
2π−arccos(2.081.81097…−0.30653…)+2πn
n=1끼우다 2π−arccos(2.081.81097…−0.30653…)+2π1
sin(θ)−0.2cos(θ)=9.87.51 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(2.081.81097…−0.30653…)+2π1sin(2π−arccos(2.081.81097…−0.30653…)+2π1)−0.2cos(2π−arccos(2.081.81097…−0.30653…)+2π1)=9.87.51
다듬다−0.96617…=0.76632…
⇒거짓
θ=arccos(−2.080.30653…+1.81097…)+2πn,θ=arccos(2.081.81097…−0.30653…)+2πn
해를 10진수 형식으로 표시θ=2.48873…+2πn,θ=1.04764…+2πn