Solution to a trigonometric system

Question

Find the solutions of the system

$$\sin a-\frac{\sqrt{3}}{3}\sin b=0$$ $$\frac{\tan 2a-2\tan a}{\tan 2b}\cdot\frac{\tan 2b-2\tan b}{\tan 2a} =1$$

How to work with them ? Thanks

Answer 1

Let's take a look at the second equation :

$\frac{\tan 2a-2\tan a}{\tan 2b}\cdot\frac{\tan 2b-2\tan b}{\tan 2a} =1$

$\frac{\tan 2a-2\tan a}{\tan 2a}\cdot\frac{\tan 2b-2\tan b}{\tan 2b} =1$

Let's take a look at :

$1-\frac{2\tan b}{\tan 2b} \implies 1-\frac{2\frac{\sin b}{\cos b}}{\frac{2\sin b \cos b}{cos^2 b - \sin^2 b}}=\tan^2 b$

$\tan^2(a) \tan^2(b) = 1$

Then $$3\sin^2 a =\sin^2 b$$ $$\tan^2(a) \tan^2(b) = 1$$

I think it's easier to see that

$\sin^2 a = \frac{1}{4}$