Question

Find all solutions in the interval [0, 2π).
tan x + sec x = 1
These are the answer choices:
(a)x=5pi/4
(b)no solution
(c)x=0
(d)x=pi/4

Answer 1

\[\frac{\sin x}{\cos x}+\frac{1}{\cos x}=1\]

Answer 2

Can you explain further I don't quite understand @Mertsj

Answer 3

tanx = sinx/cosx and
sec x = 1/cos x

Answer 4

and How can this lead to the answer? @Mertsj

Answer 5

you could do the other way...
you get plug in the results to see which work
and if none work
it must be no solution

Answer 6

alright so let me just work it out really quick and I'll ask if I got it right, that's okay? @freckles

Answer 7

though @Mertsj way will work too :)

Answer 8

Is the answer C @freckles?

Answer 9

tan(0)=0 and sec(0)=1 and 0+1=1 :)

Answer 10

I can show you the other way too

Answer 11

Alright how is it?

Answer 12

\[\frac{\sin(x)}{\cos(x)}+\frac{1}{\cos(x)}=1 \\ \frac{\sin(x)+1}{\cos(x)}=1 \\ \sin(x)+1=\cos(x) \\ 1=\cos(x)-\sin(x) \\ \text{ multiply both sides by } \frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2}=\cos(x) \frac{\sqrt{2}}{2}- \sin(x) \frac{\sqrt{2}}{2} \\ \text{ recall } \cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2} \\ \text{ so we could write our equation as } \\ \frac{\sqrt{2}}{2}=\cos(x) \cos(\frac{\pi}{4})-\sin(x) \sin(\frac{\pi}{4}) \\ \frac{\sqrt{2}}{2}=\cos(x+\frac{\pi}{4}) \text{ by \sum identity for cosine } \\ \arccos(\frac{\sqrt{2}}{2})=x+\frac{\pi}{4} \\ \frac{\pi}{4}=x+\frac{\pi}{4} \\ \frac{\pi}{4}-\frac{\pi}{4}=x \\ 0=x \\ x=0 \]
though there are more answers outside the restriction for x :)

Answer 13

Now I don't know if this is the exact way mert intended

Answer 14

but this was the way I came up with

Answer 15

he might have had an easier way

Answer 16

sin(x)+1=cos(x)
though looking at the this equation
it shouldn't be too hard to inspect that one solution could be x=0

Answer 17

since sin(0)=0 and cos(0)=1
and 0+1=1 is a true equation

Answer 18

Awesome thanks so much! @freckles