Which values of $a$ satisfy a trigonometric equation in a given interval?

[anemone]

Find all $a$ in the interval $\left(0,\,\dfrac{\pi}{2}\right)$ such that $\dfrac{\sqrt{3}-1}{\sin a}+\dfrac{\sqrt{3}+1}{\cos a}=4\sqrt{2}$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. greg1313
3. kaliprasad

Solution from MarkFL:

Spoiler

Because both $\sin(a)$ and $\cos(a)$ are non-zero on the given interval, we may multiply through by $\sin(a)\cos(a)$ to obtain:

\(\displaystyle \left(\sqrt{3}-1\right)\cos(a)+\left(\sqrt{3}+1\right)\sin(a)=4\sqrt{2}\sin(a)\cos(a)\)

Using a linear combination on the left and the double-angle identity for sine on the right, there results:

\(\displaystyle 2\sqrt{2}\sin\left(a+\frac{\pi}{12}\right)=2\sqrt{2}\sin(2a)\)

or:

\(\displaystyle \sin\left(a+\frac{\pi}{12}\right)=\sin(2a)\)

In light of the identity $\sin(\pi-x)=\sin(x)$, we have 2 cases to consider:

(i) \(\displaystyle \sin\left(a+\frac{\pi}{12}\right)=\sin(2a)\)

From this, we obtain:

\(\displaystyle a+\frac{\pi}{12}=2a\)

\(\displaystyle a=\frac{\pi}{12}\)

(ii) \(\displaystyle \sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a)\)

From this we obtain:

\(\displaystyle \pi-\left(a+\frac{\pi}{12}\right)=2a\)

\(\displaystyle 3a=\frac{11\pi}{12}\)

\(\displaystyle a=\frac{11\pi}{36}\)

Now, we must also consider (because of the periodicity of the sine function):

(iii) \(\displaystyle \sin\left(\pi-\left(a+\frac{\pi}{12}\right)\right)=\sin(2a-2\pi)\)

From this we obtain:

\(\displaystyle \pi-\left(a+\frac{\pi}{12}\right)=2a-2\pi\)

\(\displaystyle 3a=\frac{35\pi}{12}\)

\(\displaystyle a=\frac{35\pi}{36}\)

This solution is outside of the given interval.

In light of the identities $\sin(x+\pi)=-\sin(x)$ and $\sin(-x)=-\sin(x)$, we have 1 more case to consider:

(iv) \(\displaystyle \sin\left(a+\frac{\pi}{12}+\pi\right)=\sin(-2a)\)

From this, we obtain:

\(\displaystyle a+\frac{\pi}{12}+\pi=-2a\)

\(\displaystyle 3a=-\frac{13\pi}{12}\)

\(\displaystyle a=-\frac{13\pi}{36}\)

This solution is outside of the given interval.

All other solutions resulting from periodicity are outside the given interval.

Thus, the only two solutions in the given interval are:

\(\displaystyle a\in\left\{\frac{\pi}{12},\frac{11\pi}{36}\right\}\)