Answer: (a) \(x=\frac{\pi}{18}\) or \(x=\frac{5\pi}{18}\). (b)(ii) \(y=1.19\).

Answer: (a) \(x=\frac{\pi}{18}\) or \(x=\frac{5\pi}{18}\). (b)(ii) \(y=1.19\).

(a) The first three terms are \(\sin3x\), \(5\sin3x\), \(9\sin3x\), so the common difference is

\(4\sin3x\).

The sum of the first 20 terms is

\(\displaystyle S_{20}=\frac{20}{2}\left(2\sin3x+19(4\sin3x)\right)\).

So

\(390=10(78\sin3x)\).

Hence

\(\sin3x=\frac12\).

Since \(0\le x\le\frac{\pi}{2}\), we have \(0\le3x\le\frac{3\pi}{2}\). Therefore

\(3x=\frac{\pi}{6}\quad\text{or}\quad 3x=\frac{5\pi}{6}\).

Thus

\(x=\frac{\pi}{18}\quad\text{or}\quad x=\frac{5\pi}{18}\).

(b)(i) The common ratio is

\(\displaystyle r=\frac{10\cos^{2}y}{20\cos y}=\frac12\cos y\).

Since \(-1\le\cos y\le1\),

\(\displaystyle -\frac12\le r\le\frac12\).

Thus \(|r|\lt 1\), so the progression has a sum to infinity.

(b)(ii) The sum to infinity is

\(\displaystyle \frac{20\cos y}{1-\frac12\cos y}\).

Set this equal to 9:

\(\displaystyle \frac{20\cos y}{1-\frac12\cos y}=9\).

Then

\(20\cos y=9-\frac92\cos y\).

So

\(\displaystyle \frac{49}{2}\cos y=9\),

and hence

\(\displaystyle \cos y=\frac{18}{49}\).

Therefore

\(y=\cos^{-1}\left(\frac{18}{49}\right)\approx1.19\).