Solving trig questions

**Paymemoney** · January 14th 2010, 01:22 AM

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

**Prove It** · January 14th 2010, 01:51 AM

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 $1$ .

The answer is actually

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

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

What happens when you let $n = 1$ ?

**Prove It** · January 14th 2010, 01:57 AM

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

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

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

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

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

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

**Paymemoney** · January 14th 2010, 02:29 AM

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

why would you want to plus $\pi$ again?

**Prove It** · January 14th 2010, 02:34 AM

Originally Posted by Paymemoney

why would you want to plus $\pi$ again?

The period of the tangent function is $\pi$ .

**Paymemoney** · January 14th 2010, 02:42 AM

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

**Prove It** · January 14th 2010, 02:48 AM

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

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.

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