SOLUTION: Approximate, to the nearest 10l, the solutions of the equation that are in [0c, 360c). sec q = -1.116 I don't knoq what to do. Please help!

Algebra ->  Trigonometry-basics -> SOLUTION: Approximate, to the nearest 10l, the solutions of the equation that are in [0c, 360c). sec q = -1.116 I don't knoq what to do. Please help!      Log On


   



Question 1090326: Approximate, to the nearest 10l, the solutions of the equation that are in
[0c, 360c).
sec q = -1.116
I don't knoq what to do. Please help!

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
What did you mean by "to the nearest 10l" ?
Was it a typo and you meant to the nearest 10%5Eo ?
Are you allowed to use a calculator?
I believe you are asking for solutions to sec%28q%29=-1.116
in the interval %22%5B%220%5Eo%22%2C%22360%5Eo %22%29%22 .
The interval %22%5B%220%5Eo%22%2C%22360%5Eo %22%29%22 is the values between 0%5Eo (included) and 360%5Eo (not included),
so you are looking for solutions with 0%5Eo%3C=q%3C360%5Eo .
The trigonometric function is secant, which is the reciprocal of cosine, meaning
sec%28q%29=1%2Fcos%28q%29 .
The graph of the function cos%28x%29 ,
with x measured in degrees looks like the red curve below:
graph%28650%2C300%2C-50%2C500%2C-2%2C2%2Ccos%28x%2Api%2F180%29%2C-1%2C1%29 .
The graph of the function sec%28x%29 ,
with x measured in degrees looks like the red curve below:
graph%28650%2C300%2C-50%2C500%2C-2%2C2%2C1%2Fcos%28x%2Api%2F180%29%2C-1%2C1%29 .
The solutions to sec%28q%29=-1.116
in the interval %22%5B%220%5Eo%22%2C%22360%5Eo %22%29%22
are the x values for the two circled points marked below
.
Those two points are the only points on the graph with sec%28x%29=-1%2C116
and x between 0%5Eo and 360%5Eo .
Finding solutions:
sec%28q%29=-1.116
1%2Fcos%28q%29=-1.116
cos%28q%29=-1%2F1.116
cos%28q%29=approximately-0.8960573 .
Using the function inverse cosine in a calculator you can find
the solution for q that is between 0%5Eo and 180%5Eo ,
(the second quadrant solution):
q=approximately153.6446%5Eo .
So, q=approximately-153.6446%5E0 is third quadrant solution,
but if we want one between 180%5Eo and 360%5Eo ,
we need to add 360%5Eo , to get
q=approximately180%5Eo-153.6446%5Eo=206.3554%5Eo .