Bài 3.6 trang 36 SBT Đại số và giải tích 11: Giải các phương trình sau...

Giải các phương trình sau

a) (2cos x – sin x = 2)

b) (sin 5x + cos 5x =  – 1)

c) (8{cos ^4}x – 4cos 2x + sin 4x – 4 = 0)

d) ({sin ^6}x + {cos ^6}x + {1 over 2}sin 4x = 0)

Giải

a) 

(eqalign{
& 2cos x – sin x = 2 cr
& Leftrightarrow sqrt 5 left( {{2 over {sqrt 5 }}cos x – {1 over {sqrt 5 }}sin x} ight) = 2 cr} )

Kí hiệu α là góc mà (cos alpha  = {2 over {sqrt 5 }}) và ({ m{sin}}alpha  =  – {1 over {sqrt 5 }}), ta được phương trình

(eqalign{
& cos alpha cos x + sin alpha sin x = {2 over {sqrt 5 }} cr
& Leftrightarrow cos left( {x – alpha } ight) = cos alpha cr
& Leftrightarrow x – alpha = pm alpha + k2pi ,k in { m Z} cr
& Leftrightarrow left[ matrix{
x = 2alpha + k2pi ,k in Z hfill cr
x = k2pi ,k in Z hfill cr} ight. cr} )

b) 

(eqalign{
& sin 5x + cos 5x = – 1 cr
& Leftrightarrow sqrt 2 left( {{{sqrt 2 } over 2}sin 5x + {{sqrt 2 } over 2}cos 5x} ight) = – 1 cr
& Leftrightarrow cos {pi over 4}sin 5x + sin {pi over 4}cos 5x = – {{sqrt 2 } over 2} cr
& Leftrightarrow sin left( {5x + {pi over 4}} ight) = sin left( { – {pi over 4}} ight) cr
& Leftrightarrow left[ matrix{
5x + {pi over 4} = – {pi over 4} + k2pi ,k in Z hfill cr
5x + {pi over 4} = {{5pi } over 4} + k2pi ,k in Z hfill cr} ight. cr
& Leftrightarrow left[ matrix{
x = – {pi over {10}} + k{{2pi } over 5},k in Z hfill cr
x = {pi over 5} + k{{2pi } over 5},k in Z hfill cr} ight. cr} )

c) 

(eqalign{
& 8{cos ^4}x – 4cos 2x + sin 4x – 4 = 0 cr
& Leftrightarrow 8{left( {{{1 + cos 2x} over 2}} ight)^2} – 4cos 2x + sin 4x – 4 = 0 cr
& Leftrightarrow 2left( {1 + 2cos 2x + {{cos }^2}2x} ight) – 4cos 2x + sin 4x – 4 = 0 cr
& Leftrightarrow 2{cos ^2}2x + sin 4x – 2 = 0 cr
& Leftrightarrow 1 + cos 4x + sin 4x – 2 = 0 cr
& Leftrightarrow cos 4x + sin 4x = 1 cr
& Leftrightarrow sin left( {4x + {pi over 4}} ight) = sin {pi over 4} cr
& Leftrightarrow left[ matrix{
4x + {pi over 4} = {pi over 4} + k2pi ,k in Z hfill cr
4x + {pi over 4} = {{3pi } over 4} + k2pi ,k in Z hfill cr} ight. cr
& Leftrightarrow left[ matrix{
x = k{pi over 2},k in Z hfill cr
x = {pi over 8} + k{pi over 2},k in Z hfill cr} ight. cr} )

d)

(eqalign{
& {sin ^6}x + {cos ^6}x + {1 over 2}sin 4x = 0 cr
& Leftrightarrow {left( {{{sin }^2}x + {{cos }^2}x} ight)^3} – 3{sin ^2}x{cos ^2}xleft( {{{sin }^2}x + {{cos }^2}x} ight) + {1 over 2}sin 4x = 0 cr
& Leftrightarrow 1 – 3{sin ^2}x{cos ^2}x + {1 over 2}sin 4x = 0 cr
& Leftrightarrow 1 – 3{left( {{{sin 2x} over 2}} ight)^2} + {1 over 2}sin 4x = 0 cr
& Leftrightarrow 1 – {3 over 4}{sin ^2}2x + {1 over 2}sin 4x = 0 cr
& Leftrightarrow 1 – {3 over 4}.{{1 – cos 4x} over 2} + {1 over 2}sin 4x = 0 cr
& Leftrightarrow 8 – 3 + 3cos 4x + 4sin 4x = 0 cr
& Leftrightarrow 3cos 4x + 4sin 4x = – 5 cr
& Leftrightarrow {3 over 5}cos 4x + {4 over 5}sin 4x = – 1 cr} )

Kí hiệu α là cung mà (sin alpha  = {3 over 5},cos alpha  = {4 over 5}) ta được:

(eqalign{
& Leftrightarrow sin left( {4x + alpha } ight) = – 1 cr
& Leftrightarrow 4x + alpha = {{3pi } over 2},k in Z cr
& Leftrightarrow x = {{3pi } over 8} – {alpha over 4} + k{pi over 2},k in Z cr} )