1. And from that we see

$\displaystyle \tan{15^\circ} = \tan{2x}$

What do you think $\displaystyle x$ must be?

2. Originally Posted by Prove It
And from that we see

$\displaystyle \tan{15^\circ} = \tan{2x}$

What do you think $\displaystyle x$ must be?
In fact, since the tangent function is positive in the first and third quadrants, we actually have

$\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = 2 - \sqrt{3}$, where $\displaystyle n$ is an integer representing the number of times around the unit circle.

So solve $\displaystyle \tan{(\{15^\circ, 195^\circ\} + 360^\circ n)} = \tan{2x}$.

3. Originally Posted by Prove It
$\displaystyle \sec{(2x + 30^\circ)} = \sec{(2x - 60^\circ)}$

$\displaystyle \cos{(2x + 30^\circ)} = \cos{(2x - 60^\circ)}$.

Use the identity $\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$

So $\displaystyle \cos{2x}\cos{30^\circ} - \sin{2x}\sin{30^\circ} = \cos{2x}\cos{60^\circ} + \sin{2x}\sin{60^\circ}$

$\displaystyle \frac{\sqrt{3}}{2}\cos{2x} - \frac{1}{2}\sin{2x} = \frac{1}{2}\cos{2x} + \frac{\sqrt{3}}{2}\sin{2x}$

$\displaystyle \frac{\sqrt{3} - 1}{2}\cos{2x} = \frac{\sqrt{3} + 1}{2}\sin{2x}$

$\displaystyle \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \tan{2x}$

$\displaystyle \frac{(\sqrt{3} - 1)^2}{3 - 1} = \tan{2x}$

$\displaystyle \frac{3 - 2\sqrt{3} + 1}{2} = \tan{2x}$

$\displaystyle \frac{4 - 2\sqrt{3}}{2} = \tan{2x}$

$\displaystyle 2 - \sqrt{3} = \tan{2x}$

This might seem a bit off track, but if you're good with your half angle identities, you will know that

$\displaystyle \tan{\frac{\theta}{2}} = \frac{1 - \cos{\theta}}{\sin{\theta}}$.

Using this identity, what does $\displaystyle \tan{15^\circ}$ equal? (It will help you solve your equation...)
Taking it from here, tan2x=tan^-1\frac{1 - \cos{\theta}}{\sin{\theta}}
= 15

Since tan2x=positive, it is in quadrant 1 and 3

So x=15, 195, 360+15, 360+195
= 15, 195, 375, 555

4. Originally Posted by Punch
Taking it from here, tan2x=tan^-1\frac{1 - \cos{\theta}}{\sin{\theta}}
= 15

Since tan2x=positive, it is in quadrant 1 and 3

So x=15, 195, 360+15, 360+195
= 15, 195, 375, 555
No.

You have $\displaystyle \{15^\circ, 195^\circ\} + 360^\circ n = 2x$

So $\displaystyle x = \{7.5^\circ, 97.5^\circ\} + 180^\circ n$.

Also, don't try and evaluate all possibilities. Since you do not have a restricted domain, you would need an infinity of them.

Just write the answer in terms of $\displaystyle n$.

5. Oh yes, I forgot to devide it by 2 to get x values... And no, there is a domain and that is between 0 and 360

6. Originally Posted by Punch
Find all the angles between $\displaystyle 0^\circ$ and $\displaystyle 360^\circ$ which satisfy the equation $\displaystyle sec(2x+30^\circ)=sec(2x-60^\circ)$
I thought of something and that is,

does the domain of this question applies for $\displaystyle 0^\circ<x<360^\circ$ or $\displaystyle 0^\circ<2x<360^\circ$?

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