Toán Giải phương trình: cox2x+Cos^5x – sin^5x= 0 05/10/2021 By Serenity Giải phương trình: cox2x+Cos^5x – sin^5x= 0
$cos^2x – sin^2x =-(cos^5x – sin^5x)$ <->$ (cosx – sinx) (cosx + sinx) = -(cosx – sinx)( cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$ <-> $cosx -sin x =0$ hoac $cosx + sinx = -(cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$ TH1: $cosx – sinx = 0$ Vay $tanx = 1$ hay $x = \pi/2 + k\pi$. TH2: $cosx + sinx = -(cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$ <-> $cosx + sinx = -(cos^4x + sin^4x + cos^2x sin^2x + sinx cosx (sin^2x + cos^2x)$ <->$ cosx + sinx = -[(cos^2x + sin^2x)^2 – cos^2x sin^2x + sinx cosx$ <->$cosx + sinx = cos^2x sin^2x-1$ <->$ (cosx + sinx)^2 = (cos^2x sin^2x -1)^2$ <-> $ 1+2sinx cosx = sin^4x cos^4x -2sin^2x cos^2x + 1$ <-> $ sinx cosx (2-sin^3x cos^3x + 2sinx cosx) = 0$ <-> $sinx = 0$ hoac $cosx = 0$ hoac $2-sin^3x cos^3x + 2sinx cosx=0$ <-> $x = k\pi$ hoac $x = \pi/2 + k\pi$ hoac $sin^3x cos^3x – 2sinx cosx-2=0$ Xet ptrinh $sin^3x cos^3x – 2sinx cosx-2=0$ Dat $t = sinx cosx$ (ta co t = 1/2 sin 2x, vay $-1/2 \leq t \leq 1/2$) Ptrinh tro thanh $t^3 -2t-2=0$. (Ban bam may giai ptrinh thi ptrinh vo nghiem nhe) Vay $x = k \pi$ hoac $x = \pi/2 + k\pi$ Trả lời
Đáp án: Giải thích các bước giải: b.<=> sin4x=−cos5x⇔cos(4x+π2)=cos5x=> 4x + π/2 = 5x=> x= π/2c,sin 5x*cos 3x = sin 6x*cos 2x<=> sin8x + sin2x = sin8x + sin4x<=> sin2x = sin4x<=> [ 2x= 4x+k2pi[ 2x= pi – 4x +k2pid,tan5x*tan3x = 1 ( đk cosx5 khác 0 và cos3 khác 0)<=> sin5x.sin3x = cos5x.cos3x<=> cos2x – cos8x = cos8x + cos2x<=> cos8x=0 giải rồi kết hợp với đk Trả lời
$cos^2x – sin^2x =-(cos^5x – sin^5x)$
<->$ (cosx – sinx) (cosx + sinx) = -(cosx – sinx)( cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$
<-> $cosx -sin x =0$ hoac $cosx + sinx = -(cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$
TH1: $cosx – sinx = 0$
Vay $tanx = 1$ hay $x = \pi/2 + k\pi$.
TH2: $cosx + sinx = -(cos^4x + cos^3x sinx + cos^2x sin^2x + cosx sin^3x + sin^4x)$
<-> $cosx + sinx = -(cos^4x + sin^4x + cos^2x sin^2x + sinx cosx (sin^2x + cos^2x)$
<->$ cosx + sinx = -[(cos^2x + sin^2x)^2 – cos^2x sin^2x + sinx cosx$
<->$cosx + sinx = cos^2x sin^2x-1$
<->$ (cosx + sinx)^2 = (cos^2x sin^2x -1)^2$
<-> $ 1+2sinx cosx = sin^4x cos^4x -2sin^2x cos^2x + 1$
<-> $ sinx cosx (2-sin^3x cos^3x + 2sinx cosx) = 0$
<-> $sinx = 0$ hoac $cosx = 0$ hoac $2-sin^3x cos^3x + 2sinx cosx=0$
<-> $x = k\pi$ hoac $x = \pi/2 + k\pi$ hoac $sin^3x cos^3x – 2sinx cosx-2=0$
Xet ptrinh $sin^3x cos^3x – 2sinx cosx-2=0$
Dat $t = sinx cosx$ (ta co t = 1/2 sin 2x, vay $-1/2 \leq t \leq 1/2$)
Ptrinh tro thanh
$t^3 -2t-2=0$.
(Ban bam may giai ptrinh thi ptrinh vo nghiem nhe)
Vay $x = k \pi$ hoac $x = \pi/2 + k\pi$
Đáp án:
Giải thích các bước giải:
b.
<=> sin4x=−cos5x
⇔cos(4x+π2)=cos5x
=> 4x + π/2 = 5x
=> x= π/2
c,
sin 5x*cos 3x = sin 6x*cos 2x
<=> sin8x + sin2x = sin8x + sin4x
<=> sin2x = sin4x
<=> [ 2x= 4x+k2pi
[ 2x= pi – 4x +k2pi
d,
tan5x*tan3x = 1 ( đk cosx5 khác 0 và cos3 khác 0)
<=> sin5x.sin3x = cos5x.cos3x
<=> cos2x – cos8x = cos8x + cos2x
<=> cos8x=0 giải rồi kết hợp với đk