Giải các phương trình sau: a) 2 sin ( x^2 − 15^0 ) − 1 = 0 .

a) \(2\sin \left( {\frac{x}{2} - {{15}^0}} \right) - 1 = 0\)

\(\begin{array}{l} \Leftrightarrow \sin \left( {\frac{x}{2} - {{15}^0}} \right) = \frac{1}{2}\\ \Leftrightarrow \sin \left( {\frac{x}{2} - {{15}^0}} \right) = \sin {30^0}\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} - {15^0} = {30^0} + k{360^0}\\\frac{x}{2} - {15^0} = {180^0} - {30^0} + k{360^0}\end{array} \right.\;\;\left( {k \in \mathbb{Z}} \right)\end{array}\)

\( \Leftrightarrow \left[ \begin{array}{l}x = {90^0} + k{720^0}\\x = {330^0} + k{720^0}\end{array} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

Vậy phương trình có nghiệm là: \(\left[ \begin{array}{l}x = {90^0} + k{720^0}\\x = {330^0} + k{720^0}\end{array} \right.\;\;\left( {k \in \mathbb{Z}} \right)\).

b) \(2{\cos ^2}x - \sin 3x = 1\)

\(\begin{array}{l} \Leftrightarrow 2{\cos ^2}x - 1 = \sin 3x\\ \Leftrightarrow \cos 2x = \sin 3x\\ \Leftrightarrow \cos 2x = \cos \left( {\frac{\pi }{2} - 3x} \right)\end{array}\)

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

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{2} - k2\pi \end{array} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)

Vậy phương trình có nghiệm là: \(\left[ \begin{array}{l}x = \frac{\pi }{{10}} + k\frac{{2\pi }}{5}\\x = \frac{\pi }{2} - k2\pi \end{array} \right.\;\;\left( {k \in \mathbb{Z}} \right)\).