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

$\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

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

$\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

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

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

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

What happens when you let $\displaystyle n = 1$?

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

$\displaystyle cosec_(2x)=1$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Will edit.