Answer: (a) \(\displaystyle \phi=\frac\pi3,\frac\pi2,\frac{5\pi}{6}\); (b) \(\displaystyle y=\frac{1}{2x-2}-1\).
Answer: (a) \(\displaystyle \phi=\frac\pi3,\frac\pi2,\frac{5\pi}{6}\); (b) \(\displaystyle y=\frac{1}{2x-2}-1\).
(a) From
\(3\operatorname{cosec}^{2}\left(2\phi-\frac{\pi}{3}\right)=4,\)
we get
\(\operatorname{cosec}^{2}\left(2\phi-\frac{\pi}{3}\right)=\frac43.\)
Hence
\(\sin^{2}\left(2\phi-\frac{\pi}{3}\right)=\frac34.\)
Let
\(u=2\phi-\frac{\pi}{3}.\)
Since \(0\lt \phi\lt \pi\),
\(-\frac{\pi}{3}\lt u\lt \frac{5\pi}{3}.\)
In this interval, \(\sin^{2}u=\frac34\) gives
\(u=\frac\pi3,\quad\frac{2\pi}{3},\quad\frac{4\pi}{3}.\)
Therefore
\(2\phi-\frac\pi3=\frac\pi3,\quad\frac{2\pi}{3},\quad\frac{4\pi}{3}.\)
Solving for \(\phi\),
\(\phi=\frac\pi3,\quad\frac\pi2,\quad\frac{5\pi}{6}.\)
(b) Use the identity
\(\operatorname{cosec}^{2}\theta=1+\operatorname{cot}^{2}\theta.\)
Since \(2x-1=\operatorname{cosec}^{2}\theta\),
\(\operatorname{cot}^{2}\theta=2x-2.\)
Therefore
\(\tan^{2}\theta=\frac{1}{2x-2}.\)
But \(y+1=\tan^{2}\theta\), so
\(y+1=\frac{1}{2x-2}.\)
Hence
\(y=\frac{1}{2x-2}-1.\)
