# Thread: solving sine

1. ## solving sine

i've been having problems with these...

sin x= -.5 ,0 < x <2pi

sinx =-1/2

x=11pi/6 +2kpi

2. Originally Posted by cumana
i've been having problems with these...

sin x= -.5 ,0 < x <2pi

sinx =-1/2

x=11pi/6 +2kpi
Not quite. Try thinking of it this way: We get a reference angle of $\displaystyle x = \frac{\pi}{6}$.

Now, sine is negative in two quadrants, III and IV. So we have solutions for $\displaystyle sin(x) = -\frac{1}{2}$:
$\displaystyle x = \frac{7 \pi}{6}$
and
$\displaystyle x = \frac{11 \pi}{6}$

So the solution is
$\displaystyle x = \frac{7 \pi}{6} + 2k \pi$
and
$\displaystyle x = \frac{11 \pi}{6} + 2m \pi$
(You really don't need to change the arbitrary integer from k to m, but I do it to remind myself that the two do not need to be the same.)

-Dan

3. Hello, cumana!

$\displaystyle \sin x\:=\:-\frac{1}{2},\;\;{\bf{\color{red}0< x < 2\pi}}$

Since $\displaystyle \sin\frac{\pi}{6} \:=\:\frac{1}{2}$, and sine is negative in Quadrants 3 and 4,

. . then: .$\displaystyle \boxed{x \;=\;\frac{7\pi}{6},\:\frac{11\pi}{6}}$

Those are the only answers in the given interval.