# Solving a trigonometric equation

• Jul 14th 2012, 11:45 AM
Frederick
Solving a trigonometric equation
The problem:

http://s13.postimage.org/8rlczwkyv/solve.png

Hi. I was able to find the solution, option D, by plugging the different angles. I'm not sure if this is the best way to solve the problem though. Any ideas?
• Jul 14th 2012, 12:00 PM
Plato
Re: Solving a trigonometric equation
Quote:

Originally Posted by Frederick
The problem:
http://s13.postimage.org/8rlczwkyv/solve.png
Hi. I was able to find the solution, option D, by plugging the different angles. I'm not sure if this is the best way to solve the problem though. Any ideas?

You might note the it is equivalent to: $\displaystyle 2\sec^2(x)-3\sec(x)-2=0.$
• Jul 14th 2012, 12:01 PM
HallsofIvy
Re: Solving a trigonometric equation
First, do you know the basic definitions and trig identities? The very first thing I recommend is writing every thing in terms of sine and cosine, using the definitions of "tangent" and "secant". Then use a trig identity to write a quadratic equation in terms or "cosine" only and, finally, solve the equation.

I see Plato got in just ahead of me and suggested writing it as a quadratic equation in "secant" rather than "cosine"! Much of a muchness.
• Jul 14th 2012, 01:13 PM
Frederick
Re: Solving a trigonometric equation
I tried the quadratic formula way for a different problem but I didn't get the right solution:

http://s14.postimage.org/z6kdxz5n5/problem.jpg

is there a difference between writing it as a quadratic formula in sine or any of the other basic trig definitions?
• Jul 14th 2012, 01:31 PM
Plato
Re: Solving a trigonometric equation
Quote:

Originally Posted by Frederick
I tried the quadratic formula way for a different problem but I didn't get the right solution:
http://s14.postimage.org/z6kdxz5n5/problem.jpg
is there a difference between writing it as a quadratic formula in sine or any of the other basic trig definitions?

$\displaystyle (2\sec(x)+1)(\sec(x)-2)=0$ tells us that $\displaystyle sec(x)=2$ or $\displaystyle x=\frac{\pi}{3}~\&~\frac{5\pi}{3}$.
• Jul 14th 2012, 01:58 PM
Frederick
Re: Solving a trigonometric equation
I found out that the solution of the last problem I posted can be found by simply factoring. Changing to secant is unnecessary.
• Jul 14th 2012, 02:14 PM
Plato
Re: Solving a trigonometric equation