Answer: \(x=15^\circ,38.9^\circ,75^\circ,98.9^\circ\).

Use the identity

\(\sec^2\theta=1+\tan^2\theta\).

With \(\theta=3x\), the equation becomes

\(1+\tan^2 3x+\tan3x-3=0\).

So

\(\tan^2 3x+ an3x-2=0\).

Factorising,

\((\tan3x+2)(\tan3x-1)=0\).

Hence

\(\tan3x=1\quad\text{or}\quad\tan3x=-2\).

Since

\(0^\circ\leqslant x\leqslant120^\circ\),

we have

\(0^\circ\leqslant3x\leqslant360^\circ\).

For \(\tan3x=1\), the solutions in this interval are

\(3x=45^\circ,225^\circ\).

Therefore

\(x=15^\circ,75^\circ\).

For \(\tan3x=-2\), the reference angle is

\(\tan^{-1}2=63.4349\ldots^\circ\).

Since tangent is negative in quadrants II and IV,

\(3x=180^\circ-63.4349\ldots^\circ=116.565\ldots^\circ\),

or

\(3x=360^\circ-63.4349\ldots^\circ=296.565\ldots^\circ\).

Thus

\(x=38.855\ldots^\circ\quad\text{or}\quad x=98.855\ldots^\circ\).

To 3 significant figures,

\(x=15^\circ,38.9^\circ,75^\circ,98.9^\circ\).

Answer: \(x=0^\circ,38.0^\circ,90^\circ,128^\circ,180^\circ\), and \(y=-4.59,-0.947,1.69\).

(a) Factorise the equation:

\(\tan^2(2x)-4\tan(2x)=\tan(2x)(\tan(2x)-4)=0\).

So

\(\tan(2x)=0\quad\text{or}\quad \tan(2x)=4\).

Since \(0^\circ\leqslant x\leqslant180^\circ\), we have

\(0^\circ\leqslant2x\leqslant360^\circ\).

For \(\tan(2x)=0\),

\(2x=0^\circ,180^\circ,360^\circ\),

so

\(x=0^\circ,90^\circ,180^\circ\).

For \(\tan(2x)=4\),

\(2x=\tan^{-1}4=75.963\ldots^\circ\)

or, adding \(180^\circ\),

\(2x=255.963\ldots^\circ\).

Thus

\(x=37.981\ldots^\circ\quad\text{or}\quad x=127.981\ldots^\circ\).

To 3 significant figures, the solutions are

\(x=0^\circ,38.0^\circ,90^\circ,128^\circ,180^\circ\).

(b) Since

\(\operatorname{cosec}(y+1.2)=4\),

we have

\(\sin(y+1.2)=\frac14\).

Let

\(u=y+1.2\).

The given interval \(-5\lt y\lt2\) becomes

\(-3.8\lt u\lt3.2\).

The principal value is

\(\sin^{-1}\left(\frac14\right)=0.25268\ldots\).

In the interval \(-3.8\lt u\lt3.2\), the solutions are

\(u=0.25268\ldots,\quad u=\pi-0.25268\ldots,\quad u=-\pi-0.25268\ldots\).

Subtracting \(1.2\) from each value gives

\(y=-0.947\ldots,\quad y=1.688\ldots,\quad y=-4.594\ldots\).

Therefore, to 3 significant figures,

\(y=-4.59,-0.947,1.69\).

Answer: \(\theta=96.4^\circ\) or \(\theta=623.6^\circ\).

(a) Start with the left-hand side:

\(\dfrac{1-\sin x}{\cos x}+\dfrac{\cos x}{1-\sin x}.\)

Use the common denominator \(\cos x(1-\sin x)\):

\(\dfrac{(1-\sin x)^2+\cos^2x}{\cos x(1-\sin x)}.\)

Now expand the numerator:

\((1-\sin x)^2+\cos^2x=1-2\sin x+\sin^2x+\cos^2x.\)

Since \(\sin^2x+\cos^2x=1\), this becomes

\(2-2\sin x=2(1-\sin x).\)

Therefore

\(\dfrac{(1-\sin x)^2+\cos^2x}{\cos x(1-\sin x)}=\dfrac{2(1-\sin x)}{\cos x(1-\sin x)}=\dfrac2{\cos x}=2\sec x.\)

This proves the identity.

(b) Using the identity with \(x=\dfrac\theta2\), the equation becomes

\(2\sec\dfrac\theta2=3.\)

Hence

\(\sec\dfrac\theta2=\dfrac32,\)

so

\(\cos\dfrac\theta2=\dfrac23.\)

Since

\(0^\circ\leqslant\theta\leqslant720^\circ,\)

we have

\(0^\circ\leqslant\dfrac\theta2\leqslant360^\circ.\)

In this interval, \(\cos u=\dfrac23\) has two solutions:

\(u=48.189\ldots^\circ\quad\text{or}\quad u=311.810\ldots^\circ.\)

Therefore

\(\theta=2u=96.378\ldots^\circ\quad\text{or}\quad623.621\ldots^\circ.\)

To one decimal place,

\(\theta=96.4^\circ\quad\text{or}\quad\theta=623.6^\circ.\)

Answer: \(\theta=-\dfrac{\pi}{3},\,0,\,\dfrac{\pi}{3}\).

Start with

\(2\sin^3\theta=3\sin\theta\cos\theta.\)

Bring all terms to one side:

\(2\sin^3\theta-3\sin\theta\cos\theta=0.\)

Factor out \(\sin\theta\):

\(\sin\theta(2\sin^2\theta-3\cos\theta)=0.\)

Therefore either

\(\sin\theta=0\)

or

\(2\sin^2\theta-3\cos\theta=0.\)

From \(\sin\theta=0\), in the interval

\(-\dfrac\pi2\leqslant\theta\leqslant\dfrac\pi2,\)

we get

\(\theta=0.\)

For the second factor, use \(\sin^2\theta=1-\cos^2\theta\):

\(2(1-\cos^2\theta)-3\cos\theta=0.\)

Expanding,

\(2-2\cos^2\theta-3\cos\theta=0.\)

Multiply by \(-1\):

\(2\cos^2\theta+3\cos\theta-2=0.\)

Factorise:

\((2\cos\theta-1)(\cos\theta+2)=0.\)

The equation \(\cos\theta=-2\) has no solution, so

\(\cos\theta=\dfrac12.\)

In the interval \(-\dfrac\pi2\leqslant\theta\leqslant\dfrac\pi2\), this gives

\(\theta=-\dfrac\pi3\quad\text{or}\quad\theta=\dfrac\pi3.\)

Including the solution from \(\sin\theta=0\), the solutions are

\(\theta=-\dfrac\pi3,\quad0,\quad\dfrac\pi3.\)