Answer: \(\phi=-2\pi,\ -\pi,\ \pi,\ 2\pi\).

Start by writing the trigonometric functions in terms of \(\sin\), \(\cos\), \(\tan\), \(\operatorname{sec}\), or \(\operatorname{cosec}\) as appropriate, then simplify algebraically.

(a) Start with

\(\cos^4\theta-\sin^4\theta+1.\)

Use the difference of two squares:

\(\cos^4\theta-\sin^4\theta =(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta).\)

Since \(\cos^2\theta+\sin^2\theta=1\),

\(\cos^4\theta-\sin^4\theta+1 =\cos^2\theta-\sin^2\theta+1.\)

Now use \(\sin^2\theta=1-\cos^2\theta\):

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

Therefore

\(\cos^4\theta-\sin^4\theta+1=2\cos^2\theta.\)

(b) Let

\(u=\frac{\phi}{3}.\)

Using part (a), the equation becomes

\(2\cos^2u=\frac12.\)

So

\(\cos^2u=\frac14.\)

Hence

\(\cos u=\pm\frac12.\)

Since

\(-3\pi\lt\phi\lt3\pi,\)

we have

\(-\pi\lt u\lt\pi.\)

In this interval,

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

Multiplying by 3 gives

\(\phi=-2\pi,\ -\pi,\ \pi,\ 2\pi.\)