Need help with another one... tan x + sec x = 2 cos x ??

Question

Ok so I tried playing with different identities but I can't figure this one out. Can somebody help me? Thanks!

Find x, -infinity < X < infinity (again don't know how to do symbols).

Thanks, that's the answer, but where does the K come from?

Never mind, I misunderstood why the K was there. Thanks yall :].

Answer 1

tan[x] + sec[x] = 2cos[x]

sin[x]/cos[x] + 1/cos[x] = 2cos[x] ... cross multiply cos[x]

sin[x] + 1 = 2cos²[x]
sin[x] + 1 = 2(1-sin²[x])
2sin²[x] + sin[x] - 1 = 0 ... let v = sin[x]

2v² + v - 1 = 0
(2v-1)(v+1)=0
v = 1/2 or v = -1

when sin[x] = 1/2 ... x = π/6 or 5π/6 or their other coterminals... just add/subtract 2π.
x = 2kπ ± π/6, where k is any integer.

when sin[x] = -1 ... x = π + 2kπ = (2k+1)π , where k is any integer.


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Answer 2

tan x + sec x = 2 cos x
sin/cos + 1/cos = 2 cos
multiply the right side by cos/cos (multiplying by 1 is OK and cos/cos = 1)(this gives a comoon denominator of cos everywhere)
sin/cos + 1/cos = 2 cos^2/cos
get rid of the denominator (i.e., multiply both sides by cos)
sin + 1 = 2 cos^2
Move the 1 to the right side (subtract 1 from both sides)
sin = 2 cos^2 -1

From the identity sin^2 + cos^2 = 1,
we use cos^2 = 1 - sin^2

sin = 2 (1 - sin^2) - 1
sin = 2 - 2 sin^2 -1
sin = 1 - 2 sin^2
move everything to the left:
2 sin^2 + sin - 1 = 0

a quadratic. let y = sin x
2 y^2 + y - 1 = 0
y = [-1 +/- SQRT(1 +9)] / 4
y = [-1 + 3]/4 and [-1 - 3] /4
y = +1/2 and -1

y = 1/2, then
sin x = 1/2, x = pi/6 and x = 5pi/6 (30 and 150 degrees).

y = -1, then
sin x = -1, x = 3pi/2 = 9pi/6

in each case, sin x is cyclical, with a period of 2pi.

x = 2*k*pi + pi/6, and
x = 2*k*pi + 5pi/6, and
x = 2*k*pi + 9pi/6
where k can be any integer from - infinity to + infinity

or, if you want to really be fancy:

x = (4*m + 1)pi/6
where m is any integer from -inf. to +inf.

x = {... -11pi/6, -7pi/6, -3pi/6, 1pi/6, 5pi/6, 9pi/6, 13pi/6 ...}


Always check to make sure:

x = pi/6 (= 30 deg.)
tan x = 1 / √3
sec x = 2 / √3
cos x = √3 / 2
2 cos x = √3

tan x + sec x = 2 cos x
1/√3 + 2/√3 = 3/√3 = √3

Answer 3

In a problem such as this with a lot of mixed trig functions I immediately change everything to sines and cosines and it usually works
sinx/cosx + 1/cosx = 2cosx
Multiply by cox x
sinx + 1 = 2 cox^2x or sinx + 1 = 2(1 - sin^2x)
So we have:
2sin^2x + sinx -1 = 0 or (2sinx -1)(sinx +1) =0
sinx = 1/2 and x = pi/6, 5pi/6 both + or - 2npi.
sinx = -1 at 3pi/2 + or - 2npi.
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Answer 4

tan x + sec x
= sin x/ cos x + 1/cos x
= (sinx + 1)/cosx
so (1 + sinx) / cosx = 2 cosx
(1 + sinx) = 2 cos^2x = 2 (1 - sin^2x)
let u = sinx
1 + u = 2(1 - u^2)
2u^2 -2 +1 + u = 0
2u^2 + u - 1 = 0
(2u - 1)( u + 1) = 0
u = 1/2 or u = -1

u = sin x = 1/2
x = 30 deg, 150 deg
or x = pi/6 + 2n pi, 5pi/6 + 2n pi radians, n = 0, +1, -1, +2, -2,...

u = sinx = -1
x = -90 deg or 270 deg
= 3pi/2 + 2npi radians, n = 0, +1, -1, +2, -2, +3, ...

Answer 5

Please just do not be panic.
tan x + sec x = 2 cos x
this is equivalent to:
(sin x) / (cos x) + 1/ (cos x) = 2 cos x
or: sin x + 1 = 2 cos^2 x = 2 - 2sin^2 x
or: 2sin^2 x + sin x - 1 =0
This is a quadratic equation in sin x and it can be factorized as:
(2sin x - 1)(sin x + 1) = 0
The solutions to this quadratic equation are:
sin x = -1, or x = 2kπ - π/2
and sin x = 1/2, or x = 2kπ + π/6 and 2kπ + 5π/6