a)Vì \[\frac{\pi }{2} < \alpha < \pi \] nên \[\cos \alpha < 0\].

Ta có: \[{\sin ^2}\alpha + co{s^2}\alpha = 1\].

Suy ra: \[cos\alpha = - \sqrt {1 - {{\sin }^2}\alpha } = - \frac{4}{5} \Rightarrow \tan \alpha = - \frac{3}{4}\].

Vậy \[\tan \left( {\frac{\pi }{3} + \alpha } \right) = \frac{{\tan \frac{\pi }{3} + \tan \alpha }}{{1 - \tan \frac{\pi }{3}\tan \alpha }} = \frac{{48 - 25\sqrt 3 }}{{11}}\].

b) \(\cos \left( {\frac{\pi }{3} + 3x} \right) + \cos \left( {\frac{{2\pi }}{3} - 4x} \right) + \cos x = 1\)

\( \Leftrightarrow \cos \left( {\frac{\pi }{3} + 3x} \right) + \cos \left( {\frac{{2\pi }}{3} - 4x} \right) = 1 - \cos x\)

\( \Leftrightarrow 2\cos \left( {\frac{\pi }{2} - \frac{x}{2}} \right)\cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) = 2{\sin ^2}\frac{x}{2} \Leftrightarrow 2\sin \frac{x}{2}\cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) = 2{\sin ^2}\frac{x}{2}\)

\( \Leftrightarrow \sin \frac{x}{2}\left[ {\cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) - \sin \frac{x}{2}} \right] = 0 \Leftrightarrow \sin \frac{x}{2}\left[ {\cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) - \cos \left( {\frac{\pi }{2} - \frac{x}{2}} \right)} \right] = 0.\)

● \(\sin \frac{x}{2} = 0 \Leftrightarrow \frac{x}{2} = k\pi \Leftrightarrow x = k2\pi \) \(\left( {k \in \mathbb{Z}} \right)\).

● \(\cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) - \cos \left( {\frac{\pi }{2} - \frac{x}{2}} \right) = 0 \Leftrightarrow \cos \left( {\frac{{7x}}{2} - \frac{\pi }{6}} \right) = \cos \left( {\frac{\pi }{2} - \frac{x}{2}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}\frac{{7x}}{2} - \frac{\pi }{6} = \frac{\pi }{2} - \frac{x}{2} + k2\pi \\\frac{{7x}}{2} - \frac{\pi }{6} = - \left( {\frac{\pi }{2} - \frac{x}{2}} \right) + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\frac{\pi }{2}\\x = - \frac{\pi }{9} + k\frac{{2\pi }}{3}\end{array} \right.\) \(\left( {k \in \mathbb{Z}} \right)\).

Vậy phương trình đã cho có nghiệm \(x = k2\pi \); \(x = \frac{\pi }{6} + k\frac{\pi }{2}\); \(x = - \frac{\pi }{9} + k\frac{{2\pi }}{3}\), \(\left( {k \in \mathbb{Z}} \right)\).

c) Dựa vào hệ trục ta có:

\(\tan \alpha = \frac{{OM}}{{OH}} \Rightarrow OM = OH.\tan \alpha \)

Với \(\alpha = \frac{\pi }{{10}}t\) \( \Rightarrow {y_M} = 1.\tan \left( {\frac{\pi }{{10}}t} \right) = \tan \left( {\frac{\pi }{{10}}t} \right)\)

Khi \({y_N} = - 1 \Rightarrow \tan \left( {\frac{\pi }{{10}}t} \right) = - 1\)

\( \Leftrightarrow \frac{\pi }{{10}}t = \frac{{3\pi }}{4} + k\pi ,k \in \mathbb{Z}\)

\( \Leftrightarrow t = \frac{{15}}{2} + 10k,k \in \mathbb{Z}\) và \(k \ge 0\).

$Tính giá trị lượng giác \[\tan \lef (ảnh 2)$

Parabol ( P ) : y = − 2x^2 − 6x + 3 có hoành độ đỉnh là