# A q about finiding a solution for a Trig. Function

• Nov 5th 2006, 10:09 AM
A q about finiding a solution for a Trig. Function
hi evreyone,
here is the q:

Find the solutions of the equation sin θ = 1/2 if:
(a) θ is in the interval [0, 2 pi )
(b) θ is any real number
• Nov 5th 2006, 10:12 AM
ThePerfectHacker
Quote:

hi evreyone,
here is the q:

Find the solutions of the equation sin θ = 1/2 if:
(a) θ is in the interval [0, 2 pi )
(b) θ is any real number

The solutions exist in the first and second quadrand because you are trying to find positive one.

The one in the first quadrand is $\displaystyle \pi/6$ then one in the second quadrand is, $\displaystyle \pi-\pi/6=5\pi/6$
• Nov 5th 2006, 10:28 AM
i have read in a solution for a q that sin θ = 2 has no solution, why?
• Nov 5th 2006, 10:30 AM
Quick
Quote:

i have read in a solution for a q that sin θ = 2 has no solution, why?

because $\displaystyle -1<\sin\theta<1$
• Nov 5th 2006, 10:33 AM
ThePerfectHacker
Quote:

Originally Posted by Quick
because $\displaystyle -1<\sin\theta<1$

Becuase,
$\displaystyle -1\leq \sin \theta \leq 1$

If you want to be cool :cool: you can write,
$\displaystyle |\sin \theta|<1$
• Nov 5th 2006, 12:03 PM
Quick
Quote:

Originally Posted by ThePerfectHacker
Becuase,
$\displaystyle -1\leq \sin \theta \leq 1$

If you want to be cool :cool: you can write,
$\displaystyle |\sin \theta|<1$

$\displaystyle |\sin \theta|\leq 1$ :p
• Nov 5th 2006, 12:05 PM
CaptainBlack
Quote:

hi evreyone,
here is the q:

Find the solutions of the equation sin θ = 1/2 if:
(a) θ is in the interval [0, 2 pi )
(b) θ is any real number

If $\displaystyle \theta$ is a solution of $\displaystyle \sin(\theta)=1/2$ then so is $\displaystyle \theta_n = \theta+2\pi n, \ \ n=\pm 1, \ \pm 2, ...$.

So using the solutions given by ImPerfectHacker we have:

$\displaystyle \pi/6+2 \pi n, n=0,\ \pm 1,\ \pm 2,\ ...$, and $\displaystyle 5 \pi/6+2 \pi n, n=0,\ \pm 1,\ \pm 2,\ ...$ comprise all the real solution.

RonL
• Nov 5th 2006, 01:09 PM
ThePerfectHacker
Quote:

Originally Posted by Quick
$\displaystyle |\sin \theta|\leq 1$ :p

Wow, retard after a retard.