JavaScript is disabled. For a better suffer, please enable JavaScript in your internet browser before proceeding.

You are watching: Find all solutions in the interval [0, 2π). tan x + sec x = 1


Hello,I was given the equation secx + tanx = 1 and I have to resolve it (so I get a solution prefer pi/6 or something choose that). I have tried a number of times to no avail and was hoping someone below could aid. Any assist would be substantially appreciated.Thanks,Rob
*

The equation means tan²x = (1 - sec x)² = 1 - 2sec x + sec²x = 2 - 2sec x + tan²x, so sec x = 1, and so cos x = 1, which is simple to resolve.
Here"s my solution:tan(x) + sec(x) = 1sin(x)/cos(x) + 1/cos(x) = 1(sin(x) + 1)/(cos(x)) = 1cos(x) = sin(x) + 1cos(x) = (+-)sqrt(1 - cos^2(x)) + 1 (Trig identification sin^2(x) + cos^2(x) = 1, therefore sin(x) = (+-)sqrt(1 - cos^2(x)).)cos(x) - 1 = (+-)sqrt(1 - cos^2(x))cos^2(x) - 2cos(x) + 1 = 1 - cos^2(x)0 = 2cos^2(x) - 2cos(x)0 = cos(x)(cos(x) - 1)cos(x) ?= 0 and/or cos(x) - 1 ?= 0Because the original equation has cos(x) on denominator on fractivity, we have the right to say that cos(x) =/= 0, and cos(x) - 1 = 0, for this reason cos(x) = 1
, and plug this in the original equation, and we obtain sin(x) + 1 = 1, so tan(x) = 0, hence we also get sec(x) = 1.
*

Dividing sec²x - tan²x = 1 by the provided equation offers sec x - tan x = 1, and also adding that to the original equation and then halving offers sec x = 1.
Using trig identity:1 + tan^2(x) = sec^2(x)(sec(x) + tan(x))^2= 1= sec^2(x) + 2sec(x)tan(x) + tan^2(x)= 1 + tan^2(x) + tan^2(x) + 2sec(x)tan(x)0= 2tan(x)(tan(x) + sec(x))0= tan(x)(tan(x) + sec(x))sin(x)/cos(x) ?= tan(x) ?= 0 and/or tan(x) + sec(x) ?= 0If sin(x) = 0 and cos(x) =/= 0, tan(x) = 0sin(x)/cos(x) = -1/cos(x), where cos(x) =/= 0, then sin(x) = -1, however will not work bereason once sin(x) = -1, then cos(x) hregarding be 0, however cannot let be equal to zero, as adhering to arithmetic rules on denominators of fractions.Hence, we deserve to say that sin(x) = 0 and cos(x) =/= 0, then tan(x) = 0. Plug this in the original equation, and we getsec(x) + tan(x)= 1= sec(x).
*

1 = tan x + sec x = (sin x + 1)/cos x = (-cos(pi/2 + x) + 1)/sin(pi/2 + x) = 2sin²(pi/4 + x/2)/(2sin(pi/4 + x/2)cos(pi/4 + x/2)) = tan(pi/4 + x/2).Hence pi/4 + x/2 = pi/4 + kpi (where k is an integer), so x = 2kpi.
*

(displaystyle sec x+ an x=1)(displaystyle sec x+ an x=sec x- an x)(displaystyle an x=- an x)which is just true as soon as $x$ is an integer multiple of $2pi$.

See more: Cheat Engine Divinity Original Sin Enhanced Edition General Discussions

*
(displaystyle sec x+ an x=1quad<1>)(displaystyle sec x+ an x=(sec x+ an x)(sec x- an x)impliessec x- an x=1quad<2>)(displaystyle <1>+<2>Rightarrow2sec x=2impliessec x=1indicates x=2kpi,kinmathbfZ)(This strategy is used in skipjack"s write-up #4)
A couple of points to allude out.First of all, we must always think of the implied domain names of the attributes. As both $displaystyle eginalign* sec(x) endalign*$ and $displaystyle eginalign* an(x) endalign*$ are defined with department of $displaystyle eginalign* cos(x) endalign*$, that suggests that the function will not be identified where $displaystyle eginalign* cos(x) = 0 endalign*$, i.e. wright here $displaystyle eginalign* x = fracleft( 2,n + 1 ight) pi2 endalign*$, wright here $displaystyle eginalign* n in mathbfZ endalign*$.Second, at times it is unpreventable, yet if you require squaring an equation in order to settle it, you have the right to bring in extraneous options. Once you have actually got a solution, you actually have to check if they all actually fulfill the original equation.This is why I primarily like to not square equations if I deserve to avoid it.$displaystyle eginalign* sec(x) + an(x) &= 1 \ frac1cos(x) + fracsin(x)cos(x) &= 1 \ frac1 + sin(x)cos(x) &= 1 \ 1 + sin(x) &= cos(x) \ left< 1 + sin(x) ight> ^2 &= cos^2(x) \ 1 + 2sin(x) + sin^2(x) &= cos^2(x) \ 1 + 2sin(x) + sin^2(x) &= 1 - sin^2(x) \ 2sin(x) + 2sin^2(x) &= 0 \ 2sin(x)left< 1 + sin(x) ight> &= 0 endalign*$Case 1:$displaystyle eginalign* sin(x) &= 0 \ x &= m,pi extrm where m in mathbfZ endalign*$Now as I shelp, as we have squared the equation, we need to inspect whether every one of these services work-related in the original equation, so$displaystyle eginalign* sec left( m,pi ight) + an left( m,pi ight) &= frac1cos left( m,pi ight) + 0 \ &= frac1left( -1 ight) ^m \ &= left( -1 ight) ^m endalign*$So just the times as soon as m is even will certainly offer a solution to the original equation.Hence a solution to the equation is $displaystyle eginalign* 2,k,pi endalign*$ wbelow $displaystyle eginalign* k in mathbfZ endalign*$.Case 2:$displaystyle eginalign* 1 + sin(x) &= 0 \ sin(x) &= -1 \ x &= frac left( 4,p - 1 ight) pi2 extrm wbelow p in mathbfZ endalign*$But we have already established that any odd multiple of $displaystyle eginalign* fracpi2 endalign*$ is not in the domajor of the original equation, so namong these options work.So the finish solution to the equation $displaystyle eginalign* sec(x) + an(x) = 0 endalign*$ is $displaystyle eginalign* x = 2,k,pi extrm wright here k in mathbfZ endalign*$.