해법
sin(θ)−0.2cos(θ)=0.6377
해법
θ=2.66345…+2πn,θ=0.87293…+2πn
+1
도
θ=152.60452…∘+360∘n,θ=50.01533…∘+360∘n솔루션 단계
sin(θ)−0.2cos(θ)=0.6377
더하다 0.2cos(θ) 양쪽으로sin(θ)=0.6377+0.2cos(θ)
양쪽을 제곱sin2(θ)=(0.6377+0.2cos(θ))2
빼다 (0.6377+0.2cos(θ))2 양쪽에서sin2(θ)−0.40666129−0.25508cos(θ)−0.04cos2(θ)=0
삼각성을 사용하여 다시 쓰기
−0.40666129+sin2(θ)−0.04cos2(θ)−0.25508cos(θ)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.40666129+1−cos2(θ)−0.04cos2(θ)−0.25508cos(θ)
−0.40666129+1−cos2(θ)−0.04cos2(θ)−0.25508cos(θ)간소화하다 :−1.04cos2(θ)−0.25508cos(θ)+0.59333871
−0.40666129+1−cos2(θ)−0.04cos2(θ)−0.25508cos(θ)
유사 요소 추가: −cos2(θ)−0.04cos2(θ)=−1.04cos2(θ)=−0.40666129+1−1.04cos2(θ)−0.25508cos(θ)
숫자 더하기/ 빼기: −0.40666129+1=0.59333871=−1.04cos2(θ)−0.25508cos(θ)+0.59333871
=−1.04cos2(θ)−0.25508cos(θ)+0.59333871
0.59333871−0.25508cos(θ)−1.04cos2(θ)=0
대체로 해결
0.59333871−0.25508cos(θ)−1.04cos2(θ)=0
하게: cos(θ)=u0.59333871−0.25508u−1.04u2=0
0.59333871−0.25508u−1.04u2=0:u=−2.080.25508+2.53335484,u=2.082.53335484−0.25508
0.59333871−0.25508u−1.04u2=0
표준 양식으로 작성 ax2+bx+c=0−1.04u2−0.25508u+0.59333871=0
쿼드 공식으로 해결
−1.04u2−0.25508u+0.59333871=0
4차 방정식 공식:
위해서 a=−1.04,b=−0.25508,c=0.59333871u1,2=2(−1.04)−(−0.25508)±(−0.25508)2−4(−1.04)⋅0.59333871
u1,2=2(−1.04)−(−0.25508)±(−0.25508)2−4(−1.04)⋅0.59333871
(−0.25508)2−4(−1.04)⋅0.59333871=2.53335484
(−0.25508)2−4(−1.04)⋅0.59333871
규칙 적용 −(−a)=a=(−0.25508)2+4⋅1.04⋅0.59333871
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−0.25508)2=0.255082=0.255082+4⋅0.59333871⋅1.04
숫자를 곱하시오: 4⋅1.04⋅0.59333871=2.46828…=0.255082+2.46828…
0.255082=0.0650658064=0.0650658064+2.46828…
숫자 추가: 0.0650658064+2.46828…=2.53335484=2.53335484
u1,2=2(−1.04)−(−0.25508)±2.53335484
솔루션 분리u1=2(−1.04)−(−0.25508)+2.53335484,u2=2(−1.04)−(−0.25508)−2.53335484
u=2(−1.04)−(−0.25508)+2.53335484:−2.080.25508+2.53335484
2(−1.04)−(−0.25508)+2.53335484
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.25508+2.53335484
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.25508+2.53335484
분수 규칙 적용: −ba=−ba=−2.080.25508+2.53335484
u=2(−1.04)−(−0.25508)−2.53335484:2.082.53335484−0.25508
2(−1.04)−(−0.25508)−2.53335484
괄호 제거: (−a)=−a,−(−a)=a=−2⋅1.040.25508−2.53335484
숫자를 곱하시오: 2⋅1.04=2.08=−2.080.25508−2.53335484
분수 규칙 적용: −b−a=ba0.25508−2.53335484=−(2.53335484−0.25508)=2.082.53335484−0.25508
2차 방정식의 해는 다음과 같다:u=−2.080.25508+2.53335484,u=2.082.53335484−0.25508
뒤로 대체 u=cos(θ)cos(θ)=−2.080.25508+2.53335484,cos(θ)=2.082.53335484−0.25508
cos(θ)=−2.080.25508+2.53335484,cos(θ)=2.082.53335484−0.25508
cos(θ)=−2.080.25508+2.53335484:θ=arccos(−2.080.25508+2.53335484)+2πn,θ=−arccos(−2.080.25508+2.53335484)+2πn
cos(θ)=−2.080.25508+2.53335484
트리거 역속성 적용
cos(θ)=−2.080.25508+2.53335484
일반 솔루션 cos(θ)=−2.080.25508+2.53335484cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.080.25508+2.53335484)+2πn,θ=−arccos(−2.080.25508+2.53335484)+2πn
θ=arccos(−2.080.25508+2.53335484)+2πn,θ=−arccos(−2.080.25508+2.53335484)+2πn
cos(θ)=2.082.53335484−0.25508:θ=arccos(2.082.53335484−0.25508)+2πn,θ=2π−arccos(2.082.53335484−0.25508)+2πn
cos(θ)=2.082.53335484−0.25508
트리거 역속성 적용
cos(θ)=2.082.53335484−0.25508
일반 솔루션 cos(θ)=2.082.53335484−0.25508cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.082.53335484−0.25508)+2πn,θ=2π−arccos(2.082.53335484−0.25508)+2πn
θ=arccos(2.082.53335484−0.25508)+2πn,θ=2π−arccos(2.082.53335484−0.25508)+2πn
모든 솔루션 결합θ=arccos(−2.080.25508+2.53335484)+2πn,θ=−arccos(−2.080.25508+2.53335484)+2πn,θ=arccos(2.082.53335484−0.25508)+2πn,θ=2π−arccos(2.082.53335484−0.25508)+2πn
해법을 원래 방정식에 연결하여 검증
솔루션을 에 연결하여 확인합니다 sin(θ)−0.2cos(θ)=0.6377
방정식에 맞지 않는 것은 제거하십시오.
솔루션 확인 arccos(−2.080.25508+2.53335484)+2πn:참
arccos(−2.080.25508+2.53335484)+2πn
n=1끼우다 arccos(−2.080.25508+2.53335484)+2π1
sin(θ)−0.2cos(θ)=0.6377 위한 {\ quad}끼우다{\ quad} θ=arccos(−2.080.25508+2.53335484)+2π1sin(arccos(−2.080.25508+2.53335484)+2π1)−0.2cos(arccos(−2.080.25508+2.53335484)+2π1)=0.6377
다듬다0.6377=0.6377
⇒참
솔루션 확인 −arccos(−2.080.25508+2.53335484)+2πn:거짓
−arccos(−2.080.25508+2.53335484)+2πn
n=1끼우다 −arccos(−2.080.25508+2.53335484)+2π1
sin(θ)−0.2cos(θ)=0.6377 위한 {\ quad}끼우다{\ quad} θ=−arccos(−2.080.25508+2.53335484)+2π1sin(−arccos(−2.080.25508+2.53335484)+2π1)−0.2cos(−arccos(−2.080.25508+2.53335484)+2π1)=0.6377
다듬다−0.28255…=0.6377
⇒거짓
솔루션 확인 arccos(2.082.53335484−0.25508)+2πn:참
arccos(2.082.53335484−0.25508)+2πn
n=1끼우다 arccos(2.082.53335484−0.25508)+2π1
sin(θ)−0.2cos(θ)=0.6377 위한 {\ quad}끼우다{\ quad} θ=arccos(2.082.53335484−0.25508)+2π1sin(arccos(2.082.53335484−0.25508)+2π1)−0.2cos(arccos(2.082.53335484−0.25508)+2π1)=0.6377
다듬다0.6377=0.6377
⇒참
솔루션 확인 2π−arccos(2.082.53335484−0.25508)+2πn:거짓
2π−arccos(2.082.53335484−0.25508)+2πn
n=1끼우다 2π−arccos(2.082.53335484−0.25508)+2π1
sin(θ)−0.2cos(θ)=0.6377 위한 {\ quad}끼우다{\ quad} θ=2π−arccos(2.082.53335484−0.25508)+2π1sin(2π−arccos(2.082.53335484−0.25508)+2π1)−0.2cos(2π−arccos(2.082.53335484−0.25508)+2π1)=0.6377
다듬다−0.89473…=0.6377
⇒거짓
θ=arccos(−2.080.25508+2.53335484)+2πn,θ=arccos(2.082.53335484−0.25508)+2πn
해를 10진수 형식으로 표시θ=2.66345…+2πn,θ=0.87293…+2πn