해법
cos(x)−sin(2x)=cos(3x)−sin(4x)
해법
x=2π+2πn,x=23π+2πn,x=2πn,x=π+2πn,x=0.94247…+2πn,x=π−0.94247…+2πn,x=−0.31415…+2πn,x=π+0.31415…+2πn
+1
도
x=90∘+360∘n,x=270∘+360∘n,x=0∘+360∘n,x=180∘+360∘n,x=54∘+360∘n,x=126∘+360∘n,x=−18∘+360∘n,x=198∘+360∘n솔루션 단계
cos(x)−sin(2x)=cos(3x)−sin(4x)
빼다 cos(3x)−sin(4x) 양쪽에서cos(x)−sin(2x)−cos(3x)+sin(4x)=0
삼각성을 사용하여 다시 쓰기
−cos(3x)+cos(x)−sin(2x)+sin(4x)
제품식별에 대한 합계 사용: sin(s)−sin(t)=2sin(2s−t)cos(2s+t)=−cos(3x)+cos(x)+2sin(24x−2x)cos(24x+2x)
2sin(24x−2x)cos(24x+2x)=2sin(x)cos(3x)
2sin(24x−2x)cos(24x+2x)
24x−2x=x
24x−2x
유사 요소 추가: 4x−2x=2x=22x
숫자를 나눕니다: 22=1=x
=2sin(x)cos(24x+2x)
24x+2x=3x
24x+2x
유사 요소 추가: 4x+2x=6x=26x
숫자를 나눕니다: 26=3=3x
=2sin(x)cos(3x)
=−cos(3x)+cos(x)+2sin(x)cos(3x)
cos(3x)=4cos3(x)−3cos(x)
cos(3x)
삼각성을 사용하여 다시 쓰기
cos(3x)
로 고쳐 쓰다=cos(2x+x)
앵글섬식별사용: cos(s+t)=cos(s)cos(t)−sin(s)sin(t)=cos(2x)cos(x)−sin(2x)sin(x)
더블 앵글 아이덴티티 사용: sin(2x)=2sin(x)cos(x)=cos(2x)cos(x)−2sin(x)cos(x)sin(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)단순화하세요:cos(x)cos(2x)−2sin2(x)cos(x)
cos(2x)cos(x)−2sin(x)cos(x)sin(x)
2sin(x)cos(x)sin(x)=2sin2(x)cos(x)
2sin(x)cos(x)sin(x)
지수 규칙 적용: ab⋅ac=ab+csin(x)sin(x)=sin1+1(x)=2cos(x)sin1+1(x)
숫자 추가: 1+1=2=2cos(x)sin2(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
=cos(x)cos(2x)−2sin2(x)cos(x)
더블 앵글 아이덴티티 사용: cos(2x)=2cos2(x)−1=(2cos2(x)−1)cos(x)−2sin2(x)cos(x)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)확대한다:4cos3(x)−3cos(x)
(2cos2(x)−1)cos(x)−2(1−cos2(x))cos(x)
=cos(x)(2cos2(x)−1)−2cos(x)(1−cos2(x))
cos(x)(2cos2(x)−1)확대한다:2cos3(x)−cos(x)
cos(x)(2cos2(x)−1)
분배 법칙 적용: a(b−c)=ab−aca=cos(x),b=2cos2(x),c=1=cos(x)2cos2(x)−cos(x)1
=2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)−1⋅cos(x)단순화하세요:2cos3(x)−cos(x)
2cos2(x)cos(x)−1cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
지수 규칙 적용: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
숫자 추가: 2+1=3=2cos3(x)
1⋅cos(x)=cos(x)
1cos(x)
곱하다: 1⋅cos(x)=cos(x)=cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)
=2cos3(x)−cos(x)−2(1−cos2(x))cos(x)
−2cos(x)(1−cos2(x))확대한다:−2cos(x)+2cos3(x)
−2cos(x)(1−cos2(x))
분배 법칙 적용: a(b−c)=ab−aca=−2cos(x),b=1,c=cos2(x)=−2cos(x)1−(−2cos(x))cos2(x)
마이너스 플러스 규칙 적용−(−a)=a=−2⋅1cos(x)+2cos2(x)cos(x)
−2⋅1⋅cos(x)+2cos2(x)cos(x)단순화하세요:−2cos(x)+2cos3(x)
−2⋅1cos(x)+2cos2(x)cos(x)
2⋅1⋅cos(x)=2cos(x)
2⋅1cos(x)
숫자를 곱하시오: 2⋅1=2=2cos(x)
2cos2(x)cos(x)=2cos3(x)
2cos2(x)cos(x)
지수 규칙 적용: ab⋅ac=ab+ccos2(x)cos(x)=cos2+1(x)=2cos2+1(x)
숫자 추가: 2+1=3=2cos3(x)
=−2cos(x)+2cos3(x)
=−2cos(x)+2cos3(x)
=2cos3(x)−cos(x)−2cos(x)+2cos3(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)단순화하세요:4cos3(x)−3cos(x)
2cos3(x)−cos(x)−2cos(x)+2cos3(x)
집단적 용어=2cos3(x)+2cos3(x)−cos(x)−2cos(x)
유사 요소 추가: 2cos3(x)+2cos3(x)=4cos3(x)=4cos3(x)−cos(x)−2cos(x)
유사 요소 추가: −cos(x)−2cos(x)=−3cos(x)=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=4cos3(x)−3cos(x)
=−(4cos3(x)−3cos(x))+cos(x)+2(4cos3(x)−3cos(x))sin(x)
−(4cos3(x)−3cos(x))+cos(x)+2(4cos3(x)−3cos(x))sin(x)간소화하다 :−4cos3(x)+4cos(x)+8cos3(x)sin(x)−6sin(x)cos(x)
−(4cos3(x)−3cos(x))+cos(x)+2(4cos3(x)−3cos(x))sin(x)
=−(4cos3(x)−3cos(x))+cos(x)+2sin(x)(4cos3(x)−3cos(x))
−(4cos3(x)−3cos(x)):−4cos3(x)+3cos(x)
−(4cos3(x)−3cos(x))
괄호 배포=−(4cos3(x))−(−3cos(x))
마이너스 플러스 규칙 적용−(−a)=a,−(a)=−a=−4cos3(x)+3cos(x)
=−4cos3(x)+3cos(x)+cos(x)+2(4cos3(x)−3cos(x))sin(x)
2sin(x)(4cos3(x)−3cos(x))확대한다:8cos3(x)sin(x)−6sin(x)cos(x)
2sin(x)(4cos3(x)−3cos(x))
분배 법칙 적용: a(b−c)=ab−aca=2sin(x),b=4cos3(x),c=3cos(x)=2sin(x)⋅4cos3(x)−2sin(x)⋅3cos(x)
=2⋅4cos3(x)sin(x)−2⋅3sin(x)cos(x)
2⋅4cos3(x)sin(x)−2⋅3sin(x)cos(x)단순화하세요:8cos3(x)sin(x)−6sin(x)cos(x)
2⋅4cos3(x)sin(x)−2⋅3sin(x)cos(x)
숫자를 곱하시오: 2⋅4=8=8cos3(x)sin(x)−2⋅3sin(x)cos(x)
숫자를 곱하시오: 2⋅3=6=8cos3(x)sin(x)−6sin(x)cos(x)
=8cos3(x)sin(x)−6sin(x)cos(x)
=−4cos3(x)+3cos(x)+cos(x)+8cos3(x)sin(x)−6sin(x)cos(x)
유사 요소 추가: 3cos(x)+cos(x)=4cos(x)=−4cos3(x)+4cos(x)+8cos3(x)sin(x)−6sin(x)cos(x)
=−4cos3(x)+4cos(x)+8cos3(x)sin(x)−6sin(x)cos(x)
4cos(x)−4cos3(x)−6cos(x)sin(x)+8cos3(x)sin(x)=0
4cos(x)−4cos3(x)−6cos(x)sin(x)+8cos3(x)sin(x)요인:2cos(x)(2−2cos2(x)−3sin(x)+4cos2(x)sin(x))
4cos(x)−4cos3(x)−6cos(x)sin(x)+8cos3(x)sin(x)
지수 규칙 적용: ab+c=abacsin(x)cos3(x)=cos(x)cos2(x),cos3(x)=cos(x)cos2(x)=4cos(x)−4cos(x)cos2(x)−6sin(x)cos(x)+8cos(x)cos2(x)
84⋅2 로 다시 씁니다 −63⋅2 로 다시 씁니다 =2⋅2cos(x)+2⋅2cos(x)cos2(x)+3⋅2sin(x)cos(x)+2⋅2⋅2cos(x)cos2(x)
공통 용어를 추출하다 2cos(x)=2cos(x)(2−2cos2(x)−3sin(x)+4sin(x)cos2(x))
2cos(x)(2−2cos2(x)−3sin(x)+4cos2(x)sin(x))=0
각 부분을 개별적으로 해결cos(x)=0or2−2cos2(x)−3sin(x)+4cos2(x)sin(x)=0
cos(x)=0:x=2π+2πn,x=23π+2πn
cos(x)=0
일반 솔루션 cos(x)=0
cos(x) 주기율표 2πn 주기:
x06π4π3π2π32π43π65πcos(x)12322210−21−22−23xπ67π45π34π23π35π47π611πcos(x)−1−23−22−210212223
x=2π+2πn,x=23π+2πn
x=2π+2πn,x=23π+2πn
2−2cos2(x)−3sin(x)+4cos2(x)sin(x)=0:x=2πn,x=π+2πn,x=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn,x=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
2−2cos2(x)−3sin(x)+4cos2(x)sin(x)=0
삼각성을 사용하여 다시 쓰기
2−2cos2(x)−3sin(x)+4cos2(x)sin(x)
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=2−2(1−sin2(x))−3sin(x)+4(1−sin2(x))sin(x)
2−2(1−sin2(x))−3sin(x)+4(1−sin2(x))sin(x)간소화하다 :2sin2(x)+sin(x)−4sin3(x)
2−2(1−sin2(x))−3sin(x)+4(1−sin2(x))sin(x)
=2−2(1−sin2(x))−3sin(x)+4sin(x)(1−sin2(x))
−2(1−sin2(x))확대한다:−2+2sin2(x)
−2(1−sin2(x))
분배 법칙 적용: a(b−c)=ab−aca=−2,b=1,c=sin2(x)=−2⋅1−(−2)sin2(x)
마이너스 플러스 규칙 적용−(−a)=a=−2⋅1+2sin2(x)
숫자를 곱하시오: 2⋅1=2=−2+2sin2(x)
=2−2+2sin2(x)−3sin(x)+4(1−sin2(x))sin(x)
4sin(x)(1−sin2(x))확대한다:4sin(x)−4sin3(x)
4sin(x)(1−sin2(x))
분배 법칙 적용: a(b−c)=ab−aca=4sin(x),b=1,c=sin2(x)=4sin(x)⋅1−4sin(x)sin2(x)
=4⋅1⋅sin(x)−4sin2(x)sin(x)
4⋅1⋅sin(x)−4sin2(x)sin(x)단순화하세요:4sin(x)−4sin3(x)
4⋅1⋅sin(x)−4sin2(x)sin(x)
4⋅1⋅sin(x)=4sin(x)
4⋅1⋅sin(x)
숫자를 곱하시오: 4⋅1=4=4sin(x)
4sin2(x)sin(x)=4sin3(x)
4sin2(x)sin(x)
지수 규칙 적용: ab⋅ac=ab+csin2(x)sin(x)=sin2+1(x)=4sin2+1(x)
숫자 추가: 2+1=3=4sin3(x)
=4sin(x)−4sin3(x)
=4sin(x)−4sin3(x)
=2−2+2sin2(x)−3sin(x)+4sin(x)−4sin3(x)
2−2+2sin2(x)−3sin(x)+4sin(x)−4sin3(x)단순화하세요:2sin2(x)+sin(x)−4sin3(x)
2−2+2sin2(x)−3sin(x)+4sin(x)−4sin3(x)
유사 요소 추가: −3sin(x)+4sin(x)=sin(x)=2−2+2sin2(x)+sin(x)−4sin3(x)
2−2=0=2sin2(x)+sin(x)−4sin3(x)
=2sin2(x)+sin(x)−4sin3(x)
=2sin2(x)+sin(x)−4sin3(x)
sin(x)+2sin2(x)−4sin3(x)=0
대체로 해결
sin(x)+2sin2(x)−4sin3(x)=0
하게: sin(x)=uu+2u2−4u3=0
u+2u2−4u3=0:u=0,u=41+5,u=41−5
u+2u2−4u3=0
u+2u2−4u3인수 :−u(4u2−2u−1)
u+2u2−4u3
지수 규칙 적용: ab+c=abacu2=uu=−4u2u+2uu+u
공통 용어를 추출하다 −u=−u(4u2−2u−1)
−u(4u2−2u−1)=0
제로 인자 원리 사용:\4각형이면 ab=0그렇다면 a=0or b=0u=0or4u2−2u−1=0
4u2−2u−1=0해결 :u=41+5,u=41−5
4u2−2u−1=0
쿼드 공식으로 해결
4u2−2u−1=0
4차 방정식 공식:
위해서 a=4,b=−2,c=−1u1,2=2⋅4−(−2)±(−2)2−4⋅4(−1)
u1,2=2⋅4−(−2)±(−2)2−4⋅4(−1)
(−2)2−4⋅4(−1)=25
(−2)2−4⋅4(−1)
규칙 적용 −(−a)=a=(−2)2+4⋅4⋅1
지수 규칙 적용: (−a)n=an,이면 n 균등하다(−2)2=22=22+4⋅4⋅1
숫자를 곱하시오: 4⋅4⋅1=16=22+16
22=4=4+16
숫자 추가: 4+16=20=20
의 주요 인수 분해 20:22⋅5
20
20로 나누다 220=10⋅2=2⋅10
10로 나누다 210=5⋅2=2⋅2⋅5
2,5 모두 소수이므로 더 이상의 인수 분해는 불가능하다=2⋅2⋅5
=22⋅5
=22⋅5
급진적인 규칙 적용: =522
급진적인 규칙 적용: 22=2=25
u1,2=2⋅4−(−2)±25
솔루션 분리u1=2⋅4−(−2)+25,u2=2⋅4−(−2)−25
u=2⋅4−(−2)+25:41+5
2⋅4−(−2)+25
규칙 적용 −(−a)=a=2⋅42+25
숫자를 곱하시오: 2⋅4=8=82+25
2+25요인:2(1+5)
2+25
로 고쳐 쓰다=2⋅1+25
공통 용어를 추출하다 2=2(1+5)
=82(1+5)
공통 요인 취소: 2=41+5
u=2⋅4−(−2)−25:41−5
2⋅4−(−2)−25
규칙 적용 −(−a)=a=2⋅42−25
숫자를 곱하시오: 2⋅4=8=82−25
2−25요인:2(1−5)
2−25
로 고쳐 쓰다=2⋅1−25
공통 용어를 추출하다 2=2(1−5)
=82(1−5)
공통 요인 취소: 2=41−5
2차 방정식의 해는 다음과 같다:u=41+5,u=41−5
해결책은u=0,u=41+5,u=41−5
뒤로 대체 u=sin(x)sin(x)=0,sin(x)=41+5,sin(x)=41−5
sin(x)=0,sin(x)=41+5,sin(x)=41−5
sin(x)=0:x=2πn,x=π+2πn
sin(x)=0
일반 솔루션 sin(x)=0
sin(x) 주기율표 2πn 주기:
x06π4π3π2π32π43π65πsin(x)02122231232221xπ67π45π34π23π35π47π611πsin(x)0−21−22−23−1−23−22−21
x=0+2πn,x=π+2πn
x=0+2πn,x=π+2πn
x=0+2πn해결 :x=2πn
x=0+2πn
0+2πn=2πnx=2πn
x=2πn,x=π+2πn
sin(x)=41+5:x=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn
sin(x)=41+5
트리거 역속성 적용
sin(x)=41+5
일반 솔루션 sin(x)=41+5sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnx=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn
x=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn
sin(x)=41−5:x=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
sin(x)=41−5
트리거 역속성 적용
sin(x)=41−5
일반 솔루션 sin(x)=41−5sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnx=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
x=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
모든 솔루션 결합x=2πn,x=π+2πn,x=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn,x=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
모든 솔루션 결합x=2π+2πn,x=23π+2πn,x=2πn,x=π+2πn,x=arcsin(41+5)+2πn,x=π−arcsin(41+5)+2πn,x=arcsin(41−5)+2πn,x=π+arcsin(−41−5)+2πn
해를 10진수 형식으로 표시x=2π+2πn,x=23π+2πn,x=2πn,x=π+2πn,x=0.94247…+2πn,x=π−0.94247…+2πn,x=−0.31415…+2πn,x=π+0.31415…+2πn