Giải các phương trình sau
a) \(2\cos x - \sin x = 2\)
b) \(\sin 5x + \cos 5x = - 1\)
c) \(8{\cos ^4}x - 4\cos 2x + \sin 4x - 4 = 0\)
d) \({\sin ^6}x + {\cos ^6}x + {1 \over 2}\sin 4x = 0\)
a)
\(\eqalign{
& 2\cos x - \sin x = 2 \cr
& \Leftrightarrow \sqrt 5 \left( {{2 \over {\sqrt 5 }}\cos x - {1 \over {\sqrt 5 }}\sin x} \right) = 2 \cr} \)
Kí hiệu α là góc mà \(\cos \alpha = {2 \over {\sqrt 5 }}\) và \({\rm{sin}}\alpha = - {1 \over {\sqrt 5 }}\), ta được phương trình
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\(\eqalign{
& \cos \alpha \cos x + \sin \alpha \sin x = {2 \over {\sqrt 5 }} \cr
& \Leftrightarrow \cos \left( {x - \alpha } \right) = \cos \alpha \cr
& \Leftrightarrow x - \alpha = \pm \alpha + k2\pi ,k \in {\rm Z} \cr
& \Leftrightarrow \left[ \matrix{
x = 2\alpha + k2\pi ,k \in Z \hfill \cr
x = k2\pi ,k \in Z \hfill \cr} \right. \cr} \)
b)
\(\eqalign{
& \sin 5x + \cos 5x = - 1 \cr
& \Leftrightarrow \sqrt 2 \left( {{{\sqrt 2 } \over 2}\sin 5x + {{\sqrt 2 } \over 2}\cos 5x} \right) = - 1 \cr
& \Leftrightarrow \cos {\pi \over 4}\sin 5x + \sin {\pi \over 4}\cos 5x = - {{\sqrt 2 } \over 2} \cr
& \Leftrightarrow \sin \left( {5x + {\pi \over 4}} \right) = \sin \left( { - {\pi \over 4}} \right) \cr
& \Leftrightarrow \left[ \matrix{
5x + {\pi \over 4} = - {\pi \over 4} + k2\pi ,k \in Z \hfill \cr
5x + {\pi \over 4} = {{5\pi } \over 4} + k2\pi ,k \in Z \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = - {\pi \over {10}} + k{{2\pi } \over 5},k \in Z \hfill \cr
x = {\pi \over 5} + k{{2\pi } \over 5},k \in Z \hfill \cr} \right. \cr} \)
c)
\(\eqalign{
& 8{\cos ^4}x - 4\cos 2x + \sin 4x - 4 = 0 \cr
& \Leftrightarrow 8{\left( {{{1 + \cos 2x} \over 2}} \right)^2} - 4\cos 2x + \sin 4x - 4 = 0 \cr
& \Leftrightarrow 2\left( {1 + 2\cos 2x + {{\cos }^2}2x} \right) - 4\cos 2x + \sin 4x - 4 = 0 \cr
& \Leftrightarrow 2{\cos ^2}2x + \sin 4x - 2 = 0 \cr
& \Leftrightarrow 1 + \cos 4x + \sin 4x - 2 = 0 \cr
& \Leftrightarrow \cos 4x + \sin 4x = 1 \cr
& \Leftrightarrow \sin \left( {4x + {\pi \over 4}} \right) = \sin {\pi \over 4} \cr
& \Leftrightarrow \left[ \matrix{
4x + {\pi \over 4} = {\pi \over 4} + k2\pi ,k \in Z \hfill \cr
4x + {\pi \over 4} = {{3\pi } \over 4} + k2\pi ,k \in Z \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = k{\pi \over 2},k \in Z \hfill \cr
x = {\pi \over 8} + k{\pi \over 2},k \in Z \hfill \cr} \right. \cr} \)
d)
\(\eqalign{
& {\sin ^6}x + {\cos ^6}x + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} - 3{\sin ^2}x{\cos ^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow 1 - 3{\sin ^2}x{\cos ^2}x + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow 1 - 3{\left( {{{\sin 2x} \over 2}} \right)^2} + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow 1 - {3 \over 4}{\sin ^2}2x + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow 1 - {3 \over 4}.{{1 - \cos 4x} \over 2} + {1 \over 2}\sin 4x = 0 \cr
& \Leftrightarrow 8 - 3 + 3\cos 4x + 4\sin 4x = 0 \cr
& \Leftrightarrow 3\cos 4x + 4\sin 4x = - 5 \cr
& \Leftrightarrow {3 \over 5}\cos 4x + {4 \over 5}\sin 4x = - 1 \cr} \)
Kí hiệu α là cung mà \(\sin \alpha = {3 \over 5},\cos \alpha = {4 \over 5}\) ta được:
\(\eqalign{
& \Leftrightarrow \sin \left( {4x + \alpha } \right) = - 1 \cr
& \Leftrightarrow 4x + \alpha = {{3\pi } \over 2},k \in Z \cr
& \Leftrightarrow x = {{3\pi } \over 8} - {\alpha \over 4} + k{\pi \over 2},k \in Z \cr} \)