# [SOLVED] sin theta = -1/2 (quick q)

• Jun 7th 2009, 11:13 AM
brentwoodbc
[SOLVED] sin theta = -1/2 (quick q)
how do you solve this exactly? i get pi/4 and 5pi/4 by looking at a graph?

Question is solve over the set of real numbers

2sin^x-5sinx-3

I factored and got

{2sinx+1}{sinx-3}

sinx cant = 3 so

sinx = -1/2

answer is 7pi/6+2n, 11pi/6+2n n is an interger ?

If I use my calc I get the decimal answer right but how do I do this?
• Jun 7th 2009, 12:23 PM
Amer
Quote:

Originally Posted by brentwoodbc
how do you solve this exactly? i get pi/4 and 5pi/4 by looking at a graph?

Question is solve over the set of real numbers

2sin^x-5sinx-3

I factored and got

{2sinx+1}{sinx-3}

sinx cant = 3 so

sinx = -1/2

answer is 7pi/6+2n, 11pi/6+2n n is an interger ?

If I use my calc I get the decimal answer right but how do I do this?

• Jun 7th 2009, 12:36 PM
skeeter
$\displaystyle x = \frac{7\pi}{6} + 2k\pi$

$\displaystyle x = \frac{11\pi}{6} + 2k\pi$

$\displaystyle k \in \mathbb{Z}$
• Jun 7th 2009, 12:53 PM
brentwoodbc
But how do u find sin=.1/2 exactly?
• Jun 7th 2009, 01:08 PM
skeeter
Quote:

Originally Posted by brentwoodbc
But how do u find sin=.1/2 exactly?

you learn the unit circle ...

http://galileo.math.siu.edu/%7Emsull...unitcircle.gif
• Jun 7th 2009, 01:18 PM
JoanF
• Jun 7th 2009, 01:26 PM
HallsofIvy
One way to do it is to imagine an equilateral triangle. Dropping an altitude from one vertex of the triangle to the opposite side divides it into two right triangles and in both, the "opposite side" to one of the angles at that vertex is half the length of the hypotenuse: taking h as the length of the hypotenuse, $\displaystyle sin(\theta)= (1/2)h/h= 1/2= 0.5$. That angle is half one of the angles in an equilateral triangle: 180/3= 60 degrees or $\displaystyle \pi/3$ radians. Now, think about the fact that sine is an odd function to get $\displaystyle -\pi/3$ and, perhaps use the graph of y= sin(x) or the fact that sin(t) is the y coordinate on the unit circle, as skeeter shows, to see how to extend that.
• Jun 7th 2009, 02:06 PM
brentwoodbc
thanks ya I made a mistake with my special angles "trinagles" I used 30 degrees as pi/4 instead of pi/6 for some reason.

thanks.