Giải các phương trình sau:

a) sin 5x + cos 5x = – 1;

b) cos 3x – cos 5x = sin x;

c) 2 cos² x + cos 2x = 2;

d) sin⁴ x + cos⁴ x = \(\frac{1}{2}\)sin² 2x.

Lời giải

a) Ta có sin 5x + cos 5x = – 1

\( \Leftrightarrow \sqrt 2 \sin \left( {5x + \frac{\pi }{4}} \right) = - 1\)

\( \Leftrightarrow \sin \left( {5x + \frac{\pi }{4}} \right) = - \frac{1}{{\sqrt 2 }}\)

\( \Leftrightarrow \sin \left( {5x + \frac{\pi }{4}} \right) = \sin \left( { - \frac{\pi }{4}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}5x + \frac{\pi }{4} = - \frac{\pi }{4} + k2\pi \\5x + \frac{\pi }{4} = \pi - \left( { - \frac{\pi }{4}} \right) + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{5} + k\frac{{2\pi }}{5}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

b) Ta có cos 3x – cos 5x = sin x

\( \Leftrightarrow - 2\sin \frac{{3x + 5x}}{2}\sin \frac{{3x - 5x}}{2} = \sin x\)

\( \Leftrightarrow - 2\sin 4x\sin \left( { - x} \right) = \sin x\)

\( \Leftrightarrow 2\sin 4x\sin x - \sin x = 0\)

\( \Leftrightarrow \sin x\left( {2\sin 4x - 1} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin x = 0\\\sin 4x = \frac{1}{2}\end{array} \right.\).

+ Với sin x = 0 ta được x = kπ (k ∈ ℤ).

+ Với \(\sin 4x = \frac{1}{2}\)\( \Leftrightarrow \sin 4x = \sin \left( {\frac{\pi }{6}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}4x = \frac{\pi }{6} + k2\pi \\4x = \pi - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{24}} + k\frac{\pi }{2}\\x = \frac{{5\pi }}{{24}} + k\frac{\pi }{2}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\).

c) Ta có 2 cos² x + cos 2x = 2

⇔ (2 cos² x – 1) + cos 2x = 1

⇔ cos 2x + cos 2x = 1

⇔ 2cos 2x = 1

⇔ cos 2x = \(\frac{1}{2}\)

⇔ cos 2x = \(\cos \frac{\pi }{3}\)

⇔ 2x = \( \pm \frac{\pi }{3} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow x = \pm \frac{\pi }{6} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\).

d) Ta có sin⁴ x + cos⁴ x = \(\frac{1}{2}\)sin² 2x

\( \Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x = \frac{1}{2}{\sin ^2}2x\)

\( \Leftrightarrow 1 - \frac{1}{2}{\left( {2\sin x\cos x} \right)^2} = \frac{1}{2}{\sin ^2}2x\)

\( \Leftrightarrow 1 - \frac{1}{2}{\sin ^2}2x = \frac{1}{2}{\sin ^2}2x\)

\( \Leftrightarrow {\sin ^2}2x = 1\)

\( \Leftrightarrow \cos 2x = 0\) (do sin² 2x + cos² 2x = 1)

\( \Leftrightarrow 2x = \frac{\pi }{2} + k\pi \,\,\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow x = \frac{\pi }{4} + k\frac{\pi }{2}\,\,\left( {k \in \mathbb{Z}} \right)\).