Question

Solve:
2cot(x) + sec^2(x) = 0, where 0 <= x <= 2pi

Answer 1

2 cot x + sec² x = 0

Since:

cot x = 1 / tan x

sec² x = 1+ tan² x

The equation can be written as:

2 / tan x + 1 + tan² x = 0

or

tan² x + 1 + 2 / tan x = 0

Multiply both sides by tan x

tan² x ∙ tan x + 1 ∙ tan x + 2 ∙ tan x / tan x = 0

tan³ x + tan x + 2 = 0

Substitution:

tan x = u

u³ + u + 2 = 0

Since:

( u³ + u + 2 ) / ( u + 1 ) = u² - u + 2

u³ + u + 2 = ( u + 1 ) ( u² - u + 2 )

u³ + u + 2 = 0

become

( u + 1 ) ( u² - u + 2 ) = 0

The value of the equation will be equal to zero when the expressions in parentheses are equal to zero.

1 condition:

u + 1 = 0

Subtract 1 to both sides

u = - 1

tan x = u

tan x = - 1

2 condition:

u² - u + 2 = 0

The solutions are:

u = 1 / 2 - i √ 7 / 2

and

u = 1 / 2 + i √ 7 / 2

tan x = u

tan x = 1 / 2 - i √ 7 / 2

and

tan x = 1 / 2 + i √ 7 / 2

In the interval 0 ≤ x ≤ 2 π the values of the tangent are real numbers, so the conjugate complex value must be discarded.

So solution is:

tan x = - 1

In interval 0 ≤ x ≤ 2 π

tan x = - 1

for

x = 3 π / 4 rad

and

x = 7 π / 4 rad

Or in degrees:

x = 135°

and

x = 315°

Answer 2

2cot(x) + sec^2(x) = 0 , recall that 1 + tan^2 x = sec^2 x
A quick look at a Desmos graph shows solutions at 3π/4 and 7π/4 in your domain, so here goes ....

2cotx + 1 + tan^2 x = 0
2cotx + 1 + 1/cot^2 x = 0
let cotx = u
2u + 1 + 1/u^2 = 0
2u^3 + u^2 + 1 = 0
quick test shows u = -1 is a solution
2u^3 + u^2 + 1 = 0
(u+1)(2u^2 - u + 1) = 0 , the second factor has no real roots

then cotx = -1
tanx = -1 , the tangent is negative in quads II and IV
x = 135° or 315° which is 3π/4 or 7π/4

but your domain is 0 ≤ x ≤ 2π , and the period of tanx is π
so you have :
x = 3π/4, 7π/4, 11π/4, and 15π/4

Answer 3

check my end, take off the 11π/4 and 15π/4 , they are outside your domain,
so just 3π/4 and 7π/4

Answer 4

You guys are awesome! Thank You so much for all the help😊