# Trigonometric equation

1. Jun 2, 2004

### Johnny Leong

How to solve this equation:
tan 2x + sec 2x = cos x + sin x where 0<=x<=360
I solve it in this way but cannot find the right answer:
(sin 2x + 1) / cos 2x = cos x + sin x
(cos x + sin x)^2 / cos 2x = cos x + sin x
(cos x + sin x) / cos 2x = 1
sqrt(2) cos(x - 45) sec 2x = 1
sec 2x cos(x - 45) = 1 / sqrt(2)

x = 90 or 360

But the correct answers are x = 0 or 270 or 360.

2. Jun 2, 2004

### Chen

(cos x + sin x)2 equals one, how did you figure that it equals (sin 2x + 1)?

3. Jun 2, 2004

### Hurkyl

Staff Emeritus
That seems a pretty big leap; maybe if you finished doing the work on it?

You're thinking of cos2x + sin 2 x.

4. Jun 2, 2004

Thanks.

5. Jun 2, 2004

### Johnny Leong

I have solved the problem, I have made some careless mistakes above.
The solution is:
tan 2x + sec 2x = cos x + sin x where 0<=x<=360
(sin 2x + 1) / cos 2x = cos x + sin x
(cos x + sin x)^2 / cos 2x = cos x + sin x
(cos x + sin x) / cos 2x = 1
(cos x + sin x) / [(cos x + sin x)(cos x - sin x)] = 1
cos x - sin x = 1
sqrt(2) cos(x + 45) = 1
cos(x + 45) = 1 / sqrt(2)
Then x = 0 or 270 or 360