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

1. ## 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

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 $\pi/6$ then one in the second quadrand is, $\pi-\pi/6=5\pi/6$

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

i have read in a solution for a q that sin θ = 2 has no solution, why?
because $-1<\sin\theta<1$

5. Originally Posted by Quick
because $-1<\sin\theta<1$
Becuase,
$-1\leq \sin \theta \leq 1$

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

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

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

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

So using the solutions given by ImPerfectHacker we have:

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

RonL

8. Originally Posted by Quick
$|\sin \theta|\leq 1$
Wow, retard after a retard.