Solving $p\sin^{4}{\theta}-q\sin^{4}{\phi}=p$ and $p\cos^{4}{\theta}-q\cos^{4}{\phi}=q$ for $\theta$ and $\phi$

Question

Given $$p\sin^{4}{\theta}-q\sin^{4}{\phi}=p$$ and $$p\cos^{4}{\theta}-q\cos^{4}{\phi}=q$$ find $\theta$ and $\phi$.

Here is my solution (help improving it would be much appreciated):

$$p(1-2\cos^{2}{\theta}+\cos^{4}{\theta})-q(1-2\cos^{2}{\phi}+\cos^{4}{\phi})=p \tag1$$

Subtracting this from the original second equation gives $$p-2p\cos^{2}{\theta}-q+2q\cos^{2}{\phi}=p-q \tag2$$

Rearranging: $$\cos^{2}{\phi}=\frac{p}{q}\cos^{2}{\theta} \tag3$$

Which means $$\cos^{4}{\phi}=\frac{p^{2}}{q^{2}}\cos^{4}{\theta} \tag4$$

When substituted into the second original equation we get $$\left(p-\frac{p^{2}}{q}\right)\cos^{4}{\theta}=q \tag5$$

From this, $$\cos^{4}{\theta}=\frac{q^{2}}{qp-p^{2}} \tag6$$ and $$\cos^{4}{\phi}=\frac{p^{2}}{qp-p^{2}} \tag7$$

BUT... is it possible to obtain "nicer" expressions for $\theta$ and $\phi$ somehow?

Answer 1

If $p=q=0$, then any $(\theta,\phi)$ is a solution.

If $p\not=0$ and $q=0$, then $\theta=\frac{\pi}{2}+k\pi$ with any $\phi$ where $k$ is an integer.

If $q\not=0$, then there is no solution.

Proof :

If $p\not=0$ and $q=0$, then we have $\sin^4\theta=1$ and $\cos^4\theta=0$, so $\theta=\frac{\pi}{2}+k\pi$ with any $\phi$ where $k$ is an integer.

In the following, $q\not=0$.

$$p\sin^{4}{\theta}-q\sin^{4}{\phi}=p\tag1$$ $$p\cos^{4}{\theta}-q\cos^{4}{\phi}=q\tag2$$

From $(1)$, we have $$p(1-\sin^4\theta)=-q\sin^4\phi\tag3$$ From $(2)$, we have $$p\cos^4\theta=q(1+\cos^4\phi)\tag4$$

$(3)\times \cos^4\theta-(4)\times (1-\sin^4\theta)$ gives $$0=-q\sin^4\phi\cos^4\theta-q(1+\cos^4\phi)(1-\sin^4\theta)\tag5$$

Dividing the both sides of $(5)$ by $q\not=0$, we have $$-\sin^4\phi\cos^4\theta=(1+\cos^4\phi)(1-\sin^4\theta)\tag6$$

Suppose that $\sin\phi=0$. Then, since $p\cos^4\theta=2q$, we have $p\not=0$, so dividing the both sides of $p\sin^4\theta=p$ by $p$, we get $\sin\theta=\pm 1$ which implies $2q=p\cos^4\theta=0$, a contradiction.

Suppose that $\cos\theta=0$. Then, we have $\cos^4\phi=-1$ which is impossible.

So, we have $\sin\phi\cos\theta\not=0$.

Then, dividing the both sides of $(6)$ by $\sin^4\phi\cos^4\theta\not=0$, we have $$-1=\frac{1+\cos^4\phi}{\sin^4\phi}\cdot \frac{1-\sin^4\theta}{\cos^4\theta}$$ which is impossible since LHS is negative while RHS is non-negative.$\ \blacksquare$

Answer 2

We are given $$p\sin^4\theta-q\sin^4\phi=p,\qquad p\cos^4\theta-q\cos^4\phi=q.$$ Subtracting the second equation from the first, and adding it to the first, give respectively $$p(\sin^4\theta-\cos^4\theta)-q(\sin^4\phi-\cos^4\phi)=p-q,\qquad(1)\\ p(\sin^4\theta+\cos^4\theta)-q(\sin^4\phi+\cos^4\phi)=p+q.\qquad(2)$$ Since $\cos^2\theta+\sin^2\theta=\cos^2\phi+\sin^2\phi=1$, factorizing the LHS terms of eqn $1$ gives $$p(\sin^2\theta-\cos^2\theta)-q(\sin^2\phi-\cos^2\phi)=p-q,$$ which simplifies to $$p\cos^2\theta=q\cos^2\phi.\qquad(3)$$ Now eqn $2$ may be written $p(1-2\cos^2\theta\sin^2\theta)-q(1-2\cos^2\phi\sin^2\phi)=p+q$, which an application of eqn $3$ reduces to $$q\cos^2\phi\sin^2\phi=q(1+\sin^2\theta\cos^2\phi).$$ Observe that $\cos^2\phi\sin^2\phi=\frac14\sin^22\phi\leqslant\frac14$, while $1+\sin^2\theta\cos^2\phi\geqslant1$. It follows that $q=0.$ If $p$ is also zero, the given equations vanish and so have quite indefinite solutions. Otherwise, they reduce to $\sin^4\theta=1$ and $\cos^4\theta=0$. Hence $$\theta=\tfrac12\pi+k\pi\quad(k\in\Bbb Z),$$ while $\phi$ can take any value.