(i) We start with the expression \(\sin 2\alpha \sec \alpha\).
Using the identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\), we have:
\(\sin 2\alpha \sec \alpha = 2 \sin \alpha \cos \alpha \cdot \sec \alpha\)
Since \(\sec \alpha = \frac{1}{\cos \alpha}\), substitute to get:
\(2 \sin \alpha \cos \alpha \cdot \frac{1}{\cos \alpha} = 2 \sin \alpha\)
Thus, the simplified expression is \(2 \sin \alpha\).
(ii) Given the equation \(3 \cos 2\beta + 7 \cos \beta = 0\), use the identity \(\cos 2\beta = 2 \cos^2 \beta - 1\):
Substitute to get:
\(3(2 \cos^2 \beta - 1) + 7 \cos \beta = 0\)
Simplify to form a quadratic equation:
\(6 \cos^2 \beta - 3 + 7 \cos \beta = 0\)
\(6 \cos^2 \beta + 7 \cos \beta - 3 = 0\)
Solving this quadratic equation using the quadratic formula \(\cos \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = 7\), \(c = -3\):
\(\cos \beta = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}\)
\(\cos \beta = \frac{-7 \pm \sqrt{49 + 72}}{12}\)
\(\cos \beta = \frac{-7 \pm \sqrt{121}}{12}\)
\(\cos \beta = \frac{-7 \pm 11}{12}\)
This gives two solutions:
\(\cos \beta = \frac{4}{12} = \frac{1}{3}\)
\(\cos \beta = \frac{-18}{12} = -\frac{3}{2}\)
Since \(\cos \beta\) must be within \([-1, 1]\), the valid solution is \(\cos \beta = \frac{1}{3}\).