Answer: (a) \(\displaystyle y=\frac{2}{\sqrt{4x-x^2-3}}\); (b) \(\displaystyle \phi=-\frac{10\pi}{3},-\frac{4\pi}{3},\frac{2\pi}{3},\frac{8\pi}{3}\).

Reduce the equation to a standard trigonometric form, find the principal solutions, then add the correct multiples of the period to cover the required interval.

(a) From

\(2+\cos\theta=x,\)

we get

\(\cos\theta=x-2.\)

From

\(2\operatorname{cosec}\theta=y,\)

we have

\(\operatorname{cosec}\theta=\frac{y}{2},\)

so

\(\sin\theta=\frac{2}{y}.\)

Use the identity

\(\sin^2\theta+\cos^2\theta=1.\)

This gives

\(\left(\frac{2}{y}\right)^2+(x-2)^2=1.\)

So

\(\frac{4}{y^2}=1-(x-2)^2.\)

Therefore

\(y^2=\frac{4}{1-(x-2)^2}.\)

Since \(y\gt 2\), take the positive square root:

\(y=\frac{2}{\sqrt{1-(x-2)^2}}.\)

Equivalently,

\(y=\frac{2}{\sqrt{4x-x^2-3}}.\)

(b) Let

\(u=\frac{\phi}{2}.\)

The equation becomes

\(3\cos u=\sqrt3\sin u.\)

So

\(\tan u=\frac{3}{\sqrt3}=\sqrt3.\)

Therefore

\(u=\frac{\pi}{3}+n\pi,\)

where \(n\) is an integer. Hence

\(\phi=\frac{2\pi}{3}+2n\pi.\)

Now require

\(-4\pi\lt \phi\lt 4\pi.\)

The values that satisfy this are

Answer: \(\displaystyle x=\frac{11\pi}{24}\) or \(\displaystyle x=\frac{23\pi}{24}\).

Answer: \(\displaystyle x=\frac{11\pi}{24}\) or \(\displaystyle x=\frac{23\pi}{24}\).

Let

\(u=2x+\frac{\pi}{4}.\)

The equation becomes

\(3\sin u=\sqrt3\cos u.\)

Since \(\cos u\neq0\) for the solutions of this equation, divide by \(\cos u\):

\(3\tan u=\sqrt3.\)

Therefore

\(\tan u=\frac{1}{\sqrt3}.\)

As \(0\leq x\leq\pi\),

\(\frac{\pi}{4}\leq u\leq\frac{9\pi}{4}.\)

In this interval,

\(u=\frac{7\pi}{6}\quad\text{or}\quad u=\frac{13\pi}{6}.\)

Hence

\(2x+\frac{\pi}{4}=\frac{7\pi}{6}\)

or

\(2x+\frac{\pi}{4}=\frac{13\pi}{6}.\)

Solving gives

\(x=\frac{11\pi}{24}\quad\text{or}\quad x=\frac{23\pi}{24}.\)

Answer: \(x=-1.24,\ -0.195,\ 0.852\) to three significant figures.

Answer: \(x=-1.24,\ -0.195,\ 0.852\) to three significant figures.

Let

\(u=3x+1.2.\)

The equation becomes

\(\sqrt2\cos u=2\sin u.\)

Divide by \(\cos u\):

\(\sqrt2=2\tan u.\)

Therefore

\(\tan u=\frac{1}{\sqrt2}.\)

Now

\(-1.5\leq x\leq1.5,\)

so

\(-3.3\leq u\leq5.7.\)

The principal value is

\(\arctan\frac1{\sqrt2}\approx0.615.\)

Within the interval \(-3.3\leq u\leq5.7\), the possible values are

\(u\approx -2.526,\quad 0.615,\quad 3.757.\)

Since \(u=3x+1.2\),

\(x=\frac{u-1.2}{3}.\)

This gives

\(x\approx -1.24,\quad -0.195,\quad 0.852.\)

Answer: \(x=-\frac{7\pi}{12},-\frac{\pi}{12},\frac{5\pi}{12},\frac{11\pi}{12}\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

The equation is

\(\operatorname{cot}\left(2x+\frac{\pi}{3}\right)=\sqrt3.\)

Therefore

This occurs when

\(2x+\frac{\pi}{3}=\frac{\pi}{6}+k\pi,\)

where \(k\) is an integer. Hence

\(2x=-\frac{\pi}{6}+k\pi\)

and

\(x=-\frac{\pi}{12}+\frac{k\pi}{2}.\)

For \(-\pi\lt x\lt \pi\), the possible values are obtained from \(k=-1,0,1,2\):

Answer: \(x=0,\frac{\pi}{3},\frac{\pi}{2},\frac{5\pi}{6}\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

Since

\(\operatorname{cot}^2\left(2x-\frac{\pi}{3}\right)=\frac13,\)

we have

\(\tan^2\left(2x-\frac{\pi}{3}\right)=3.\)

Therefore

\(\tan\left(2x-\frac{\pi}{3}\right)=\pm\sqrt3.\)

Let

\(u=2x-\frac{\pi}{3}.\)

Since \(0\leq x\lt \pi\),

\(-\frac{\pi}{3}\leq u\lt \frac{5\pi}{3}.\)

In this interval, \(\tan u=\pm\sqrt3\) when

So

Adding \(\frac{\pi}{3}\) and dividing by \(2\),

Answer: \(x=-95^\circ,\ 175^\circ\).

Rearrange the equation:

\(\sqrt2\sin(x+50^\circ)=-1.\)

Thus

\(\sin(x+50^\circ)=-\frac1{\sqrt2}.\)

Let \(u=x+50^\circ\). Since \(-180^\circ\leq x\leq180^\circ\),

\(-130^\circ\leq u\leq230^\circ.\)

In this interval, \(\sin u=-\frac1{\sqrt2}\) when

\(u=-45^\circ\quad\text{or}\quad u=225^\circ.\)

Therefore

\(x=-95^\circ\quad\text{or}\quad x=175^\circ.\)