How do you find the exact value of (sin2theta)/(1+cos2theta)=4 in the interval 0<=theta<2pi?

2 Answers
Nov 27, 2016

theta~=0,42pi or theta~=1,42pi

Explanation:

Since

sin2theta=2sinthetacostheta

and

cos2theta=cos^2theta-sin^2theta

you can substitute in the give equation and get:

(2sinthetacostheta)/(1+cos^2theta-sin^2theta)=4

Then, since 1-sin^2theta=cos^2theta,

you can substitute and get:

(2sinthetacostheta)/(cos^2theta+cos^2theta)=4

(cancel2sinthetacancelcostheta)/(cancel2cos^cancel2theta)=4

tantheta=4

theta=arctan4

theta~=0,42pi or theta~=1,42pi

Nov 27, 2016

two solutions
theta_1=tan^(-1)(1/4), theta_2=tan^(-1)(1/4)+pi,

Explanation:

sin2theta=2sinthetacostheta
cos2theta=cos^2theta-sin^2theta

By replacing them into the equation we get
2sinthetacostheta=4+4cos^2theta-4sin^2theta

besides in order to make the equation homogenous we can replace 4 with 4sin^2theta+4cos^2theta so that the equation becomes

2sinthetacostheta=4cos^2theta+4sin^2theta+4cos^2theta-4sin^2theta
ruling out the opposite terms we get

8cos^2theta-2sinthetacostheta=0

and by collecting the common factor

2costheta(4costheta -sintheta)=0

It is satisfied when costheta=0 (for theta=pi/2+kpi)

and when sintheta=4costheta
that by dividing both sides for costheta (provided it is different from 0). It turns into
tantheta=1/4
that is satisfied for theta=tan^(-1)(1/4)+kpi

the other two possible solutions found for the first factor (theta=pi/2, theta=3/2pi) can not be taken as they break the existence condition for the given fractional equation