Solving Trigonometric Equations

[junior6790]

I'm having trouble getting started on a particular trig homework problem:

Directions:Find all solutions of the equation in the interval [0, 2pie] algebraically.
The problem is:

sec x + tan x=1

Any help would be much appreciated!

 

[stapel]

One method, frequently useful, is to convert everything to sines and cosines, and see where you can go from there.

Note: A "pie" is a dessert pastry; "pi" is the number approximately equal to 3.14159.

Eliz.

 

[opusman]

change your sec x to 1/cos x and tan x to sin x/cos x.

you then have a common denominator of cos x, add the numerators, you will have to eliminate from your solution the values that make cos x = 0

square both sides of equation, this will add extraneous roots you need to check for at the end.

(1+sin x)^2 / cos^2 x = 1

use a trig identity to replace cos^2 with 1 - sin^2 and factor the denominator

a (1+sin x) will cancel leaving you with (1+sinx)/(1-sin x) = 1

multiply both sides by (1-sinx) and solve,

remember to check for extraneous root with original equation

Hopefully, this helps.

 

[Guest]

Essentially you are left with

2sin x = 1 therefore sinx = 1/2
x must be 30 degrees, 150 degrees {0,360} domain

 

[skeeter]

Re: Essentially you are left with

secx + tanx = 1

1/cosx + sinx/cosx = 1

(1 + sinx)/cosx = 1

1 + sinx = cosx

1 + 2sinx + sin²x = cos²x

1 + 2sinx + sin²x = 1 - sin²x

2sin²x + 2sinx = 0

2sinx(sinx + 1) = 0

sinx = 0 ... x = 0, x = pi, and x = 2pi
only two of these solutions are valid ... check them in the original equation.

sinx = -1 ... x = 3pi/2
this solution is extraneous (invalid).