1. ## Solving trig questions

Need help on finding out what i have done wrong:
1)Solve cosec(2x)+1=2
This is what i have done

$cosec_(2x)=1$

$2x=\frac{\pi}{2},\frac{3\pi}{2}$

$x=\frac{\pi}{4} and \frac{3\pi}{4}$

Answer says $\frac{\pi}{4}$and $\frac{5\pi}{4}$

2)Solve $cot(2x-\frac{\pi}{3})=-1$

$2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}$

$2x=\frac{13\pi}{12},\frac{25\pi}{12}$

$x=\frac{13\pi}{24},\frac{25\pi}{24}$

P.S

2. Originally Posted by Paymemoney
Need help on finding out what i have done wrong:
1)Solve cosec(2x)+1=2
This is what i have done

$cosec_(2x)=1$

$2x=\frac{\pi}{2},\frac{3\pi}{2}$

$x=\frac{\pi}{4} and \frac{3\pi}{4}$

Answer says $\frac{\pi}{4}$and $\frac{5\pi}{4}$

2)Solve $cot(2x-\frac{\pi}{3})=-1$

$2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}$

$2x=\frac{13\pi}{12},\frac{25\pi}{12}$

$x=\frac{13\pi}{24},\frac{25\pi}{24}$

P.S
1) $\sin{\frac{3\pi}{2}} = -1$, not $1$.

$2x = \frac{\pi}{2} + 2\pi n$

$x = \frac{\pi}{4} + \pi n$.

What happens when you let $n = 1$?

3. Originally Posted by Paymemoney
Need help on finding out what i have done wrong:
1)Solve cosec(2x)+1=2
This is what i have done

$cosec_(2x)=1$

$2x=\frac{\pi}{2},\frac{3\pi}{2}$

$x=\frac{\pi}{4} and \frac{3\pi}{4}$

Answer says $\frac{\pi}{4}$and $\frac{5\pi}{4}$

2)Solve $cot(2x-\frac{\pi}{3})=-1$

$2x=\frac{3\pi}{4}+\frac{\pi}{3}, \frac{7\pi}{4}+\frac{\pi}{3}$

$2x=\frac{13\pi}{12},\frac{25\pi}{12}$

$x=\frac{13\pi}{24},\frac{25\pi}{24}$

P.S
$\cot{\left(2x - \frac{\pi}{3}\right)} = -1$

$\tan{\left(2x - \frac{\pi}{3}\right)} = -1$

$2x - \frac{\pi}{3} = \left\{ \frac{3\pi}{4}, \frac{7\pi}{4} \right\} + \pi n$

$2x = \left \{ \frac{13\pi}{12}, \frac{25\pi}{12} \right\} + \pi n$

$x = \left \{ \frac{13\pi}{24}, \frac{25\pi}{24} \right \} + \frac{\pi n}{2}$.

4. $
2x = \left \{ \frac{13\pi}{12}, \frac{25\pi}{12} \right\} + \pi n
$
why would you want to plus $\pi$ again?

5. Originally Posted by Paymemoney
why would you want to plus $\pi$ again?
The period of the tangent function is $\pi$.

6. so is this including tan=1 as well, coz is that how you got $\frac{\pi}{4}=\frac{\pi}{24}$

7. $\frac{\pi}{4}\neq \frac{\pi}{24}$.

I went from $\frac{25\pi}{12}$ to $\frac{\pi}{12}$ because you have gone around the unit circle once.

Though I may have made a mistake in doing that too early...

Will edit.