# Thread: Solving Trig Question

1. ## Solving Trig Question

Hi
I am having trouble solving the following equation:
$\displaystyle tan(3x-\frac{\pi}{6})$
range $\displaystyle 0,2\pi$
this is what a have done, someone tell me where my mistake is.
$\displaystyle 3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}$
$\displaystyle 3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23 \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6 },\frac{35\pi}{6}+\frac{\pi}{6}$
$\displaystyle x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36 },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3 6}$

P.S

2. Originally Posted by Paymemoney
Hi
I am having trouble solving the following equation:
$\displaystyle tan(3x-\frac{\pi}{6})$
range $\displaystyle 0,2\pi$
this is what a have done, someone tell me where my mistake is.
$\displaystyle 3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}$
$\displaystyle 3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23 \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6 },\frac{35\pi}{6}+\frac{\pi}{6}$
$\displaystyle x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36 },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3 6}$

P.S
HI

That's not an equation , maybe you missed the = ??

3. Originally Posted by Paymemoney
Hi
I am having trouble solving the following equation:
$\displaystyle tan(3x-\frac{\pi}{6})$
range $\displaystyle 0,2\pi$
this is what a have done, someone tell me where my mistake is.
$\displaystyle 3x-\frac{\pi}{6}=\pi-\frac{\pi}{6},2\pi-\frac{\pi}{6},3\pi-\frac{\pi}{6},4\pi-\frac{\pi}{6},5\pi-\frac{\pi}{6},6\pi-\frac{\pi}{6}$
$\displaystyle 3x=\frac{5\pi}{6}+\frac{\pi}{6},\frac{11\pi}{6}+\f rac{\pi}{6},\frac{17\pi}{6}+\frac{\pi}{6},\frac{23 \pi}{6}+\frac{\pi}{6},\frac{29\pi}{6}+\frac{\pi}{6 },\frac{35\pi}{6}+\frac{\pi}{6}$
$\displaystyle x=\frac{6\pi}{36},\frac{12\pi}{36},\frac{18\pi}{36 },\frac{24\pi}{36},\frac{30\pi}{36},\frac{26\pi}{3 6}$

P.S
It is meant to be: $\displaystyle tan(3x-\frac{\pi}{6})=-1$

4. Originally Posted by Paymemoney
It is meant to be: $\displaystyle tan(3x-\frac{\pi}{6})=-1$
HI

i will put it in degrees so that its easier for you to see but you will need to change it back to radian s.

tan (3x-30)=-1

3x-30 = 135 , 315 , 495 , 675 , 855 , 1035

x=55 , 115 , 175 , 235 , 295 , 355

the ones in red are the basic angles , the reference angle is 45 , since tan is -ve , it would be in the 2nd and 4th quadrant , then just add 360 for the subsequent angles . Just remember that

0<x<360 , 0<3x<1080

5. Originally Posted by Paymemoney
It is meant to be: $\displaystyle tan(3x-\frac{\pi}{6})=-1$
Tan is -1 in the 2nd and 4th quad so

$\displaystyle 3x-\frac{\pi}{6} = \frac{3\pi}{4}+2\pi n$

and

$\displaystyle 3x-\frac{\pi}{6} = \frac{7\pi}{4} +2\pi n$

6. oh yeh wrong quadant