MHB 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:

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

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

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

or:

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

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

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

From this, we obtain:

$$a+\frac{\pi}{12}=2a$$

$$a=\frac{\pi}{12}$$

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

From this we obtain:

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

$$3a=\frac{11\pi}{12}$$

$$a=\frac{11\pi}{36}$$

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

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

From this we obtain:

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

$$3a=\frac{35\pi}{12}$$

$$a=\frac{35\pi}{36}$$

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

From this, we obtain:

$$a+\frac{\pi}{12}+\pi=-2a$$

$$3a=-\frac{13\pi}{12}$$

$$a=-\frac{13\pi}{36}$$

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:

$$a\in\left\{\frac{\pi}{12},\frac{11\pi}{36}\right\}$$