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 $x = \pi/2 + k\pi$

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