sec^2 x + tan x = 0?

Question

I've gone through half a page trying to solve for x and all I'm doing now is going in circles. I have -cos^2 x = tan x now. Any help or idea would be greatly appreciated.

Answer 1

sec^2 x = 1 + tan^2 x
so write your problem as
tan^2 x + tan x +1 =0 and now it is a quadratic if you let y = tan x
so write
y^2 + y + 1 = 0
and u can use the quadratic formula which will lead to an imaginary solution which means there is no real solution and u could try graphing which will show you that sec^2 x + tan x is never equal to zero

Answer 2

Let me just reaffirm that this is sec^2 x + tan x = 0 yes?

then sec^2 x = tan^2 x +1
which gives us a quadratic eq.
tan^2 x + tan x + 1 = 0
then by completing the square with tan x = y,
y^2 + y = -1
y^2 + y + (1/2)^2 = -3/4
=> (y+ 1/2)^2 = -3/4
then y= -1/2 +/- v3i/2 = tan x
then for x use
arctan y = i/2 * ln [(1-iy)/(1+iy)]

Answer 3

sec^2 x + tan x = 0

but sec^2 x = tan^2 x + 1

so:

tan^2 x + 1 + tan x = 0

tan^2 x + tan x + 1 = 0

Answer 4

Write this as

1/cos^2(x) + sin(x)/cos(x) = 0

This is the same as

1/cos^2(x) + sin(x)cos(x)/cos^2(x) = 0

or

(1+ sin(x)cos(x))/cos^2(x) = 0

This procedure won't work if cos(x) = 0, or x = pi/2. Otherwise, we have:

1+sin(x)cos(x) = 0

Use the sum formula for sin(2x) = sin(x+x) and compare the result with the sin(x)cos(x) term and see what you can conclude. (hint: sines and cosines are always between -1 and 1)

Answer 5

sec(x)^2 + tan(x) = 0
sec(x)^2 = tan(x)^2 + 1
tan(x)^2 + 1 + tan(x) = 0
tan(x)^2 + tan(x) + 1 = 0
thank of this like this
x^2 + x + 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-1 ± sqrt(1 - 4(1)(1)))/(2(1))
x = (-1 ± sqrt(1 - 4))/2
x = (-1 ± sqrt(-3))/2
x = (-1 ± isqrt(3))/2

so this gives you

tan(x) = (-1 ± isqrt(3))/2
x = tan^-1((-1 ± isqrt(3))/2)

Answer 6

if -cos^2x = tan x, therefore, cos^2 x = -tan x

Substitute this to the problem

-tan x + tanx = 0
0 = 0

ok?!!!!

Answer 7

sec^2x+tanx=0
1+tan^2x+tanx=0
tanx=[-1+/-(1-4)^1/2]/2
imaginary roots