giải phương trình:

a) 2sin2x + sinx = 0;

b) sinx + cos3x = 0;

c) sinx + 2cosx = 0;

d) 2sin² 3x = 1;

e) cos2x = 2cosx – 1.

a) 2sin2x + sinx = 0

⇔ 4sinx.cosx + sinx = 0

⇔ sinx(4cosx + 1) = 0

\( \Leftrightarrow \left[ \begin{array}{l}{\mathop{\rm s}\nolimits} {\rm{inx}} = 0\\{\rm{cosx = }}\frac{{ - 1}}{4}\end{array} \right.\)

 \( \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = k\pi \\{\rm{x = ar}}cc{\rm{os }}\frac{{ - 1}}{4} + k2\pi \\x = - {\rm{ar}}cc{\rm{os }}\frac{{ - 1}}{4} + k2\pi \end{array} \right.\) (k ∈ ℤ)

b) sinx + cos3x = 0

\[ \Leftrightarrow sinx + sin\left( {\frac{\pi }{2} - 3x} \right) = 0\]

\[ \Leftrightarrow sinx = sin\left( {3x--\frac{\pi }{2}} \right)\]

\( \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = 3{\rm{x}} - \frac{\pi }{2} + k2\pi \\{\rm{x = }}\pi - 3{\rm{x}} - \frac{\pi }{2} + k2\pi {\rm{ }}\end{array} \right.\)(k ∈ ℤ)

\( \Leftrightarrow \left[ \begin{array}{l} - 2{\rm{x}} = - \frac{\pi }{2} + k2\pi \\{\rm{4x = }}\frac{{3\pi }}{2} + k2\pi {\rm{ }}\end{array} \right.\)(k ∈ ℤ)

\( \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = \frac{\pi }{4} + k\pi \\{\rm{x = }}\frac{{3\pi }}{8} + \frac{{k\pi }}{2}{\rm{ }}\end{array} \right.\)(k ∈ ℤ)

c) sinx + 2cosx = 0

⇔ sinx = – 2cosx

\( \Leftrightarrow \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{{\rm{cosx}}}} = \frac{{ - 2co{\rm{sx}}}}{{\cos x}}\)

⇔ tanx = – 2

⇔ x = arctan(– 2) + kπ (k ∈ ℤ)

d) 2sin² 3x = 1

\[ \Leftrightarrow si{n^2}3x = \frac{1}{2}\]

\( \Leftrightarrow \left[ \begin{array}{l}\sin 3{\rm{x}} = \frac{1}{{\sqrt 2 }}\\\sin 3{\rm{x}} = - \frac{1}{{\sqrt 2 }}\end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}\sin 3{\rm{x}} = \sin \frac{\pi }{4}\\\sin 3{\rm{x}} = \sin \frac{{ - \pi }}{4}\end{array} \right.\)

 \( \Leftrightarrow \left[ \begin{array}{l}3{\rm{x}} = \frac{\pi }{4} + k2\pi \\3{\rm{x}} = \pi - \frac{\pi }{4} + k2\pi \\3{\rm{x}} = \frac{{ - \pi }}{4} + k2\pi \\3{\rm{x}} = \pi + \frac{\pi }{4} + k2\pi \end{array} \right.\) \( \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = \frac{\pi }{{12}} + \frac{{k2\pi }}{3}\\{\rm{x}} = \frac{\pi }{4} + \frac{{k2\pi }}{3}\\{\rm{x}} = \frac{{ - \pi }}{{12}} + \frac{{k2\pi }}{3}\\{\rm{x}} = \frac{{5\pi }}{{12}} + \frac{{k2\pi }}{3}\end{array} \right.\) (k ∈ ℤ)

e) cos2x = 2cosx – 1

⇔ 2cos²x – 1 = 2cosx – 1

⇔ 2cos²x – 2cosx = 0

⇔ 2cosx(cosx – 1) = 0

\( \Leftrightarrow \left[ \begin{array}{l}{\rm{cosx = 0}}\\{\rm{cosx = 1}}\end{array} \right.\)

\[ \Leftrightarrow \left[ \begin{array}{l}{\rm{x}} = \frac{\pi }{2} + k\pi \\{\rm{x = }}k2\pi {\rm{ }}\end{array} \right.\] (k ∈ ℤ)