Answer: (a) \(x=30^\circ,150^\circ,191.5^\circ,348.5^\circ\). (b) \(y=0.361,\dfrac{\pi}{2},2.78\) radians.

(a) Use \(\cos^2x=1-\sin^2x\):

\(10(1-\sin^2x)+3\sin x=9.\)

Simplify:

\(10-10\sin^2x+3\sin x=9.\)

So

\(10\sin^2x-3\sin x-1=0.\)

Factorise:

\((5\sin x+1)(2\sin x-1)=0.\)

Hence

\(\sin x=\frac12 \quad\text{or}\quad \sin x=-\frac15.\)

For \(0^\circ\lt x\lt360^\circ\), \(\sin x=\frac12\) gives

\(x=30^\circ,\ 150^\circ.\)

Also, \(\sin x=-\frac15\) gives the third and fourth quadrant solutions

\(x=191.5^\circ,\ 348.5^\circ.\)

(b) Write \(\tan2y=\frac{\sin2y}{\cos2y}\):

\(3\frac{\sin2y}{\cos2y}=4\sin2y.\)

Multiply by \(\cos2y\):

\(3\sin2y=4\sin2y\cos2y.\)

Bring all terms to one side and factorise:

\(\sin2y(3-4\cos2y)=0.\)

So either

\(\sin2y=0 \quad\text{or}\quad \cos2y=\frac34.\)

Since \(0\lt y\lt\pi\), we have \(0\lt2y\lt2\pi\).

From \(\sin2y=0\), the valid solution is

\(2y=\pi,\qquad y=\frac{\pi}{2}.\)

From \(\cos2y=\frac34\),

\(2y=\cos^{-1}\frac34 \quad\text{or}\quad 2y=2\pi-\cos^{-1}\frac34.\)

Thus

\(y=0.361\ldots \quad\text{or}\quad y=2.780\ldots.\)

Therefore

\(y=0.361,\ \frac{\pi}{2},\ 2.78.\)