Baøi 133 : Giaûi phöông trình

CH??NG VII

PH??NG TR?NH L???NG GIA?C CH??A CA?N VA? PH??NG TR?NH L???NG GIA?C CH??A GIA? TR? TUYE?T ?O?I

A) PH??NG TR?NH L???NG GIA?C CH??A CA?N

Ca?ch gia?i :

A?p du?ng ca?c co?ng th??c

A=

B

A A

=

0 B

B A

0 =B

B 0

A

=

B

A

=

B2

Ghi chu? : Do theo ph??ng tr?nh ch?nh ly? ?a? bo? pha?n ba?t ph??ng tr?nh l???ng gia?c ne?n ta x?? ly? ?ie?u kie?n B 0 ba?ng ph??ng pha?p th?? la?i va? chu?ng to?i bo?

ca?c ba?i toa?n qua? ph??c ta?p.

Ba?i 138 : Gia?i ph??ng tr?nh 5 cos x - cos 2x + 2 sin x = 0 (*)

(*) 5 cos x - cos 2x = -2 sin x

sin x 0 5 cos x - cos 2x = 4 sin2 x

( ) ( )

sin x 0 5 cos x -

2 cos2

x

-1

=4

1 - cos2 x

sin x 0 2 cos2 x + 5 cos x - 3 = 0

sin cos

x x

=

0 1 2

cos

x

=

-3 ( loa?i )

sin x 0

x

=

?

3

+

k2, k

x

=

-

3

+

k2,

k

Ba?i 139 : Gia?i ph??ng tr?nh sin3 x + cos3 x + sin3 x cot gx + cos3 xtgx = 2 sin 2x

?ie?u kie?n :

cos sin sin

x0 x0 2x 0

sin sin

2x 2x

0 0

sin

2x

>

0

Lu?c ?o? :

(*) sin3 x + cos3 x + sin2 x cos x + cos2 x sin x = 2 sin 2x

sin2 x (sin x + cos x) + cos2 x (cos x + sin x) = 2sin 2x

( ) (sin x + cos x) sin2 x + cos2 x = 2 sin 2x

sin x + cos x 0

(sin x + cos x)2 =

2 sin

2x

2

sin

x

+

4

0

sin

x

+

4

0

1 + sin 2x = 2 sin 2x sin 2x = 1(nha?n do sin 2x > 0)

sin

x

+

4

0

sin

x

+

4

0

x

=

4

+

k, k

x

=

4

+

m2

x

=

5 4

+

m2 (loa?i) , m

x

=

4

+

m2, m

Ba?i 140 : Gia?i ph??ng tr?nh

1+

8 sin 2x. cos2

2x

=

2

sin

3x

+

4

(*)

Ta

co?

:

(*)

sin

3x

+

4

0

1 + 8 sin 2x cos2 2x

=

4

sin2

3x

+

4

sin

3x

+

4

0

1 + 4 sin 2x (1 + cos 4x)

=

2 1

- cos( 6x

+

2

)

sin

3x

+

4

0

1 + 4 sin 2x + 2 (sin 6x - sin 2x) = 2 (1 + sin 6x)

sin

3x

+

4

0

sin

3x

+

4

0

sin

2x

=

1 2

x

=

12

+

k

x

=

5 12

+

k,

k

So

la?i

v??i

?ie?u

kie?n

sin

3x

+

4

0

?Khi x = + k th? 12

sin

3x

+

4

=

sin

2

+

3k

=

cos

k

1 , (ne?u k cha?n) (nha?n)

=

-1

,

(ne?u

k

le?) (loa?i)

? Khi x = 5 + k th? 12

sin

3x

+

4

=

sin

3 2

+

3k

=

sin

-

2

+

k

-1 , ne?u k cha?n (loa?i)

=

1 ,

ne?u

k

le?

(nha?n)

Do

?o?

(*)

x

=

12

+

m2

x

=

5 12

+

(2m

+

1)

, m

Ba?i 141 : Gia?i ph??ng tr?nh 1 - sin 2x + 1 + sin 2x = 4 cos x (*)

sin x

Lu?c ?o? : (*) 1 - sin 2x + 1 + sin 2x = 2sin 2x

( hie?n nhie?n sinx = 0 kho?ng la? nghie?m , v? sinx =0 th? VT = 2, VP = 0 )

2

+

2

1 - sin2 2x

= 4 sin2 2x

sin 2x 0

1 - sin2 2x = 2 sin2 2x - 1

sin 2x 0

1 - sin2 2x = 4 sin4 2x - 4 sin2 2x + 1

sin2

2x

1 2

sin 2x 0

( ) sin2 2x 4 sin2 2x - 3 = 0

sin 2x

1

2

sin

2x

=

3 sin 2x = - 3

2

2

sin 2x

2 2

sin 2x = 3 2

2x

=

3

+

k2

2x

=

2 3

+

k2, k

x

=

6

+

k

x

=

3

+

k, k

Chu? y? : Co? the? ??a ve? ph??ng tr?nh ch??a gia? tr? tuye?t ?o?i

(*)

sin x 0

cos

x

-

sin

x

+

cos x + sin x

=

2 sin 2x

cos x - sin x + cos x + sin x = 2 sin 2x

Ba?i 142 : Gia?i ph??ng tr?nh sin x + 3 cos x + sin x + 3 cos x = 2 (*)

?a?t t = sin x +

sin

3

cos

x

=

sin

x

+

cos

3

cos

x

3

t

=

1 cos

sin

x

+

3

=

2

sin

x

+

3

3

(*) tha?nh t + t = 2

t =2-t

2 - t 0

t 2

t

=

4

-

4t

+

t2

t2

- 5t

+

4=0

t t

=

2 1

t

=

4

t

=

1

Do ?o? (*)

sin

x

+

3

=

1 2

x

+

3

=

6

+

k2

hay

x

+

3

=

5 6

+

k2, k

x

=

-

6

+

k2

x

=

2

+

k2, k

Ba?i 143 : Gia?i ph??ng tr?nh

3 tgx + 1 (sin x + 2 cos x) = 5 (sin x + 3 cos x) (*)

Chia hai ve? cu?a (*) cho cos x 0 ta ????c

(*) 3 tgx + 1 (tgx + 2) = 5(tgx + 3)

?a?t u = tgx + 1 v??i u 0 Th? u2 - 1 = tgx

( ) ( ) (*) tha?nh 3u u2 + 1 = 5 u2 + 2

3u3 - 5u2 + 3u - 10 = 0

(u - 2) (3u2 + u + 5) = 0

u = 2 3u2 + u + 5 = 0 ( vo? nghie?m)

Do ?o? (*) tgx + 1 = 2

tgx + 1 = 4

tgx

=

3

=

tg

v??i

-

2

<

<

2

x =

+

k , k

( ) Ba?i 144 : Gia?i ph??ng tr?nh 1 - cos x + cos x cos 2x = 1 sin 4x (*) 2 ( ) (*) 1 - cos x + cos x cos 2x = sin 2x cos 2x

cos x 0 cos 2x = 0

hay

1 - cos x +

cos x = sin 2x

cos x 0

2x

=

2

+

k,

k

hay

cos x 0

sin

2x

0

1 + 2 ( 1 - cos x)cosx = sin2 2x

cos x =

x 4

0 +k

2

,

k

hay

cos x 0 sin 2x 0 1 + 2 ( 1 - cos x)cosx = sin2 2x ( VT 1 VP )

cos x 0

cos x 0

x

=

?

4

+

h

hay

x=?

5 4

+

h,

h

hay

sin 2x 0 sin2 2x = 1

(1 - cos x ) cos x = 0

x

=

?

4

+

h,

h

sin 2x = 1

sin 2x = 1

hay cos x = 0 ( sin 2x = 0 ) hay cos x = 1 ( sin x = 0 sin 2x = 0 )

x

=

?

4

+

h,

h

Ba?i 145 : Gia?i ph??ng tr?nh sin3 x (1 + cot gx) + cos3 x (1 + tgx) = 2 sin x cos x (*)

(*)

sin3

x

sin

x+ sin

cos x

x

+

cos3

x

cos

x+ cos

sin x

x

=

2

sin x cos x

( ) (sin x + cos x) sin2 x + cos2 x = 2 sin x cos x

sin x + cos x 0 1 + sin 2x = 2 sin 2x

sin sin

x + cos 2x = 1

x

0

sin x =

x

+

4

+ k, 4

k

0

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