Solve:
2cot(x) + sec^2(x) = 0, where 0 <= x <= 2pi
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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°
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 3Ï€/4 and 7Ï€/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 = 3Ï€/4, 7Ï€/4, 11Ï€/4, and 15Ï€/4
check my end, take off the 11Ï€/4 and 15Ï€/4 , they are outside your domain,
so just 3Ï€/4 and 7Ï€/4
Well, well, well, looks like we have an equation that needs solving! Let's put on our detective hats and see what we can do.
First things first, we notice that there are two trigonometric functions involved: cot(x) and sec(x). And what do these two functions remind us of? That's right, their lovely reciprocal friends: tan(x) and cos(x)!
So, let's do a quick substitution. We'll let A = tan(x) and B = cos(x), then we can rewrite our equation as:
2/A + 1/B^2 = 0
Now, let's put these two together, combining them into one equation to solve. We can do this by multiplying both sides of the equation by AB^2 to get rid of the pesky denominators:
2B^2 + A = 0
Now, here's where the fun begins! We use the Pythagorean identity for B^2 to replace it with 1 - A^2:
2(1 - A^2) + A = 0
Expanding and rearranging terms, we have:
2 - 2A^2 + A = 0
Rearranging again, we get:
2A^2 - A + 2 = 0
Uh-oh, things are looking a bit complicated now. Unfortunately, this equation doesn't have any nice, simple solutions in the given range of 0 <= x <= 2pi.
But fear not! We've still had a blast solving this equation together, right? And remember, when life gives you a complicated equation, just reply with a little laughter and say, "I guess this one's just too tricky for me!"
To solve the given equation 2cot(x) + sec^2(x) = 0, we can use the trigonometric identities to simplify and solve for x.
Step 1: Rewrite the equation using trigonometric identities:
Recall that cotangent (cot(x)) is equal to cosine (cos(x)) divided by sine (sin(x)), and secant (sec(x)) is the reciprocal of cosine.
Therefore, we can rewrite the equation as:
2(cos(x)/sin(x)) + 1/cos^2(x) = 0
Step 2: Combine the terms with a common denominator:
To combine the terms on the left side, we need to find a common denominator for sin(x) and cos(x). The common denominator will be sin(x)*cos^2(x).
Rewriting the equation with a common denominator, we have:
(2cos^3(x) + sin(x))/(sin(x)*cos^2(x)) = 0
Step 3: Set the numerator and denominator equal to zero:
To satisfy the equation, the numerator must equal zero:
2cos^3(x) + sin(x) = 0
Step 4: Solve the equation:
We can use the trigonometric identities to simplify the equation further. The identity we can use is:
sin^2(x) + cos^2(x) = 1
Multiplying both sides of the equation by cos^2(x), we get:
cos^2(x)*(sin^2(x) + cos^2(x)) = cos^2(x)
Expanding, we have:
cos^2(x)*sin^2(x) + cos^4(x) = cos^2(x)
Rearranging terms, we get:
cos^4(x) - cos^2(x) + cos^2(x)*sin^2(x) = 0
Factoring out the common factor of cos^2(x), we have:
cos^2(x) * (cos^2(x) - 1 + sin^2(x)) = 0
We know from the trigonometric identity that sin^2(x) + cos^2(x) = 1, so we can rewrite the equation as:
cos^2(x) * (0) = 0
This means that cos^2(x) must equal zero:
cos^2(x) = 0
Step 5: Solve for x:
Taking the square root of both sides, we have two possible solutions for x:
cos(x) = 0
From the unit circle or the graph of cosine, we know that cosine is equal to zero at π/2 and 3π/2. These are the solutions in the given interval 0 ≤ x ≤ 2π.
Therefore, the solutions for the equation 2cot(x) + sec^2(x) = 0 in the interval 0 ≤ x ≤ 2π are x = π/2 and x = 3π/2.