해법
d=4⋅cos2(x)sin4(x)−cos2(x)+5
해법
x=arcsin2−1−4d+16d2+24d−15+2πn,x=π+arcsin−2−1−4d+16d2+24d−15+2πn,x=arcsin−2−1−4d+16d2+24d−15+2πn,x=π+arcsin2−1−4d+16d2+24d−15+2πn,x=arcsin2−1−4d−16d2+24d−15+2πn,x=π+arcsin−2−1−4d−16d2+24d−15+2πn,x=arcsin−2−1−4d−16d2+24d−15+2πn,x=π+arcsin2−1−4d−16d2+24d−15+2πn
솔루션 단계
d=4cos2(x)sin4(x)−cos2(x)+5
측면 전환4cos2(x)sin4(x)−cos2(x)+5=d
빼다 d 양쪽에서4cos2(x)sin4(x)−cos2(x)+5−d=0
4cos2(x)sin4(x)−cos2(x)+5−d단순화하세요:4cos2(x)sin4(x)−cos2(x)+5−4dcos2(x)
4cos2(x)sin4(x)−cos2(x)+5−d
요소를 분수로 변환: d=4cos2(x)d4cos2(x)=4cos2(x)sin4(x)−cos2(x)+5−4cos2(x)d⋅4cos2(x)
분모가 같기 때문에, 분수를 합친다: ca±cb=ca±b=4cos2(x)sin4(x)−cos2(x)+5−d⋅4cos2(x)
4cos2(x)sin4(x)−cos2(x)+5−4dcos2(x)=0
g(x)f(x)=0⇒f(x)=0sin4(x)−cos2(x)+5−4dcos2(x)=0
삼각성을 사용하여 다시 쓰기
5−cos2(x)+sin4(x)−4cos2(x)d
피타고라스 정체성 사용: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=5−(1−sin2(x))+sin4(x)−4(1−sin2(x))d
5−(1−sin2(x))+sin4(x)−4(1−sin2(x))d간소화하다 :sin4(x)+sin2(x)+4dsin2(x)+4−4d
5−(1−sin2(x))+sin4(x)−4(1−sin2(x))d
=5−(1−sin2(x))+sin4(x)−4d(1−sin2(x))
−(1−sin2(x)):−1+sin2(x)
−(1−sin2(x))
괄호 배포=−(1)−(−sin2(x))
마이너스 플러스 규칙 적용−(−a)=a,−(a)=−a=−1+sin2(x)
=5−1+sin2(x)+sin4(x)−4(1−sin2(x))d
−4d(1−sin2(x))확대한다:−4d+4dsin2(x)
−4d(1−sin2(x))
분배 법칙 적용: a(b−c)=ab−aca=−4d,b=1,c=sin2(x)=−4d⋅1−(−4d)sin2(x)
마이너스 플러스 규칙 적용−(−a)=a=−4⋅1⋅d+4dsin2(x)
숫자를 곱하시오: 4⋅1=4=−4d+4dsin2(x)
=5−1+sin2(x)+sin4(x)−4d+4dsin2(x)
숫자를 빼세요: 5−1=4=sin4(x)+sin2(x)+4dsin2(x)+4−4d
=sin4(x)+sin2(x)+4dsin2(x)+4−4d
4+sin2(x)+sin4(x)−4d+4sin2(x)d=0
대체로 해결
4+sin2(x)+sin4(x)−4d+4sin2(x)d=0
하게: sin(x)=u4+u2+u4−4d+4u2d=0
4+u2+u4−4d+4u2d=0:u=2−1−4d+16d2+24d−15,u=−2−1−4d+16d2+24d−15,u=2−1−4d−16d2+24d−15,u=−2−1−4d−16d2+24d−15
4+u2+u4−4d+4u2d=0
표준 양식으로 작성 anxn+…+a1x+d=0u4+(1+4d)u2+4−4d=0
다음으로 방정식 다시 쓰기 v=u2 그리고 v2=u4v2+(1+4d)v+4−4d=0
v2+(1+4d)v+4−4d=0해결 :v=2−1−4d+16d2+24d−15,v=2−1−4d−16d2+24d−15
v2+(1+4d)v+4−4d=0
쿼드 공식으로 해결
v2+(1+4d)v+4−4d=0
4차 방정식 공식:
위해서 a=1,b=1+4d,c=4−4dv1,2=2⋅1−(1+4d)±(1+4d)2−4⋅1⋅(4−4d)
v1,2=2⋅1−(1+4d)±(1+4d)2−4⋅1⋅(4−4d)
(1+4d)2−4⋅1⋅(4−4d)단순화하세요:16d2+24d−15
(1+4d)2−4⋅1⋅(4−4d)
숫자를 곱하시오: 4⋅1=4=(4d+1)2−4(−4d+4)
(1+4d)2−4(4−4d)확대한다:16d2+24d−15
(1+4d)2−4(4−4d)
(1+4d)2:1+8d+16d2
완벽한 정사각형 공식 적용: (a+b)2=a2+2ab+b2a=1,b=4d
=12+2⋅1⋅4d+(4d)2
12+2⋅1⋅4d+(4d)2단순화하세요:1+8d+16d2
12+2⋅1⋅4d+(4d)2
규칙 적용 1a=112=1=1+2⋅1⋅4d+(4d)2
2⋅1⋅4d=8d
2⋅1⋅4d
숫자를 곱하시오: 2⋅1⋅4=8=8d
(4d)2=16d2
(4d)2
지수 규칙 적용: (a⋅b)n=anbn=42d2
42=16=16d2
=1+8d+16d2
=1+8d+16d2
=1+8d+16d2−4(4−4d)
−4(4−4d)확대한다:−16+16d
−4(4−4d)
분배 법칙 적용: a(b−c)=ab−aca=−4,b=4,c=4d=−4⋅4−(−4)⋅4d
마이너스 플러스 규칙 적용−(−a)=a=−4⋅4+4⋅4d
숫자를 곱하시오: 4⋅4=16=−16+16d
=1+8d+16d2−16+16d
1+8d+16d2−16+16d단순화하세요:16d2+24d−15
1+8d+16d2−16+16d
집단적 용어=16d2+8d+16d+1−16
유사 요소 추가: 8d+16d=24d=16d2+24d+1−16
숫자 더하기/ 빼기: 1−16=−15=16d2+24d−15
=16d2+24d−15
=16d2+24d−15
v1,2=2⋅1−(1+4d)±16d2+24d−15
솔루션 분리v1=2⋅1−(1+4d)+16d2+24d−15,v2=2⋅1−(1+4d)−16d2+24d−15
v=2⋅1−(1+4d)+16d2+24d−15:2−1−4d+16d2+24d−15
2⋅1−(1+4d)+16d2+24d−15
숫자를 곱하시오: 2⋅1=2=2−(4d+1)+16d2+24d−15
−(1+4d):−1−4d
−(1+4d)
괄호 배포=−(1)−(4d)
마이너스 플러스 규칙 적용+(−a)=−a=−1−4d
=2−1−4d+16d2+24d−15
v=2⋅1−(1+4d)−16d2+24d−15:2−1−4d−16d2+24d−15
2⋅1−(1+4d)−16d2+24d−15
숫자를 곱하시오: 2⋅1=2=2−(4d+1)−16d2+24d−15
−(1+4d):−1−4d
−(1+4d)
괄호 배포=−(1)−(4d)
마이너스 플러스 규칙 적용+(−a)=−a=−1−4d
=2−1−4d−16d2+24d−15
2차 방정식의 해는 다음과 같다:v=2−1−4d+16d2+24d−15,v=2−1−4d−16d2+24d−15
v=2−1−4d+16d2+24d−15,v=2−1−4d−16d2+24d−15
다시 대체 v=u2,을 해결하다 u
u2=2−1−4d+16d2+24d−15해결 :u=2−1−4d+16d2+24d−15,u=−2−1−4d+16d2+24d−15
u2=2−1−4d+16d2+24d−15
위해서 x2=f(a) 해결책은 x=f(a),−f(a)
u=2−1−4d+16d2+24d−15,u=−2−1−4d+16d2+24d−15
u2=2−1−4d−16d2+24d−15해결 :u=2−1−4d−16d2+24d−15,u=−2−1−4d−16d2+24d−15
u2=2−1−4d−16d2+24d−15
위해서 x2=f(a) 해결책은 x=f(a),−f(a)
u=2−1−4d−16d2+24d−15,u=−2−1−4d−16d2+24d−15
해결책은
u=2−1−4d+16d2+24d−15,u=−2−1−4d+16d2+24d−15,u=2−1−4d−16d2+24d−15,u=−2−1−4d−16d2+24d−15
뒤로 대체 u=sin(x)sin(x)=2−1−4d+16d2+24d−15,sin(x)=−2−1−4d+16d2+24d−15,sin(x)=2−1−4d−16d2+24d−15,sin(x)=−2−1−4d−16d2+24d−15
sin(x)=2−1−4d+16d2+24d−15,sin(x)=−2−1−4d+16d2+24d−15,sin(x)=2−1−4d−16d2+24d−15,sin(x)=−2−1−4d−16d2+24d−15
sin(x)=2−1−4d+16d2+24d−15:x=arcsin2−1−4d+16d2+24d−15+2πn,x=π+arcsin−2−1−4d+16d2+24d−15+2πn
sin(x)=2−1−4d+16d2+24d−15
트리거 역속성 적용
sin(x)=2−1−4d+16d2+24d−15
일반 솔루션 sin(x)=2−1−4d+16d2+24d−15sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin2−1−4d+16d2+24d−15+2πn,x=π+arcsin−2−1−4d+16d2+24d−15+2πn
x=arcsin2−1−4d+16d2+24d−15+2πn,x=π+arcsin−2−1−4d+16d2+24d−15+2πn
sin(x)=−2−1−4d+16d2+24d−15:x=arcsin−2−1−4d+16d2+24d−15+2πn,x=π+arcsin2−1−4d+16d2+24d−15+2πn
sin(x)=−2−1−4d+16d2+24d−15
트리거 역속성 적용
sin(x)=−2−1−4d+16d2+24d−15
일반 솔루션 sin(x)=−2−1−4d+16d2+24d−15sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin−2−1−4d+16d2+24d−15+2πn,x=π+arcsin2−1−4d+16d2+24d−15+2πn
x=arcsin−2−1−4d+16d2+24d−15+2πn,x=π+arcsin2−1−4d+16d2+24d−15+2πn
sin(x)=2−1−4d−16d2+24d−15:x=arcsin2−1−4d−16d2+24d−15+2πn,x=π+arcsin−2−1−4d−16d2+24d−15+2πn
sin(x)=2−1−4d−16d2+24d−15
트리거 역속성 적용
sin(x)=2−1−4d−16d2+24d−15
일반 솔루션 sin(x)=2−1−4d−16d2+24d−15sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin2−1−4d−16d2+24d−15+2πn,x=π+arcsin−2−1−4d−16d2+24d−15+2πn
x=arcsin2−1−4d−16d2+24d−15+2πn,x=π+arcsin−2−1−4d−16d2+24d−15+2πn
sin(x)=−2−1−4d−16d2+24d−15:x=arcsin−2−1−4d−16d2+24d−15+2πn,x=π+arcsin2−1−4d−16d2+24d−15+2πn
sin(x)=−2−1−4d−16d2+24d−15
트리거 역속성 적용
sin(x)=−2−1−4d−16d2+24d−15
일반 솔루션 sin(x)=−2−1−4d−16d2+24d−15sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πnx=arcsin−2−1−4d−16d2+24d−15+2πn,x=π+arcsin2−1−4d−16d2+24d−15+2πn
x=arcsin−2−1−4d−16d2+24d−15+2πn,x=π+arcsin2−1−4d−16d2+24d−15+2πn
모든 솔루션 결합x=arcsin2−1−4d+16d2+24d−15+2πn,x=π+arcsin−2−1−4d+16d2+24d−15+2πn,x=arcsin−2−1−4d+16d2+24d−15+2πn,x=π+arcsin2−1−4d+16d2+24d−15+2πn,x=arcsin2−1−4d−16d2+24d−15+2πn,x=π+arcsin−2−1−4d−16d2+24d−15+2πn,x=arcsin−2−1−4d−16d2+24d−15+2πn,x=π+arcsin2−1−4d−16d2+24d−15+2πn