a) Cho sin α = 1/2 với pi/2 < α <pi. Tính sin 2α b) Giải phương trình lượng giác sau: cos 3x + cos x − sin 2x = 0

a) Cho \[{\rm{sin}}\alpha  = \frac{1}{2}\] với \[\frac{\pi }{2} < \alpha  < \pi \]. Tính \[\sin 2\alpha \]

Vì \(\frac{\pi }{2} < \alpha  < \pi \)\( \Rightarrow co{\mathop{\rm s}\nolimits} \alpha  < 0\), \(\cos \alpha  =  - \sqrt {1 - \frac{1}{4}}  = \frac{{ - \sqrt 3 }}{2}\)

sin2\(\alpha \)=\[2\sin \alpha .\cos \alpha  = 2.\left( {\frac{{ - \sqrt 3 }}{2}} \right).\left( {\frac{1}{2}} \right) =  - \frac{{\sqrt 3 }}{2}\]

b) \(\cos 3x + \cos x - \sin 2x = 0\)

\(\begin{array}{l}Pt \Leftrightarrow 2\cos 2x\cos x - 2\sin x\cos x = 0\\ \Leftrightarrow 2\cos x\left( {\cos 2x - \sin x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos 2x = \sin x = \cos \left( {\frac{\pi }{2} - x} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\\left[ \begin{array}{l}x = \frac{\pi }{6} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{2} + k2\pi \end{array} \right.\\\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k\pi \\x = \frac{\pi }{6} + k\frac{{2\pi }}{3}\end{array} \right.\end{array}\)