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°
Solve:
2cot(x) + sec^2(x) = 0, where 0 <= x <= 2pi
4 answers
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
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
check my end, take off the 11π/4 and 15π/4 , they are outside your domain,
so just 3π/4 and 7π/4
so just 3π/4 and 7π/4
You guys are awesome! Thank You so much for all the help😊