How do I solve 3 sin x+2 tan x=0 for 0<=x<=360 ?

1 Answer
Sep 28, 2015

x=0,
x=180,
x=360,
x=arccos(-2/3),
x=-arccos(-2/3)

Explanation:

The first step is to determine the domain of the functions involved. With the restriction 0<=x<=360 (degrees, I assume) we have exclude the points where tan(x) is not defined.
By definition, tan(x)=sin(x)/cos(x)
Therefore, we have to exclude points where cos(x)=0, that is the domain is described as
x != 90 and x != 270

With the above restrictions in mind we can transform the equation as follows:
3sin(x)+2sin(x)/cos(x) = 0

The immediate temptation to reduce the above equation by sin(x) should be accompanied by checking if it can be equal to zero to avoid losing solutions.
Indeed, sin(x)=0 can occur within a domain defined above.
It happens when
x=0, x=180 and x=360.
So the three values above are solutions.

After specifying these three solutions we can reduce the equation by sin(x) getting
3+2/cos(x)=0,
from which follows
cos(x)=-2/3,
solutions of this are
x = arccos(-2/3) and x = -arccos(-2/3)

CHECKING

The first three solutions (0, 180 and 360) cause both sin(x) and tan(x) to be equal to 0, therefore the left side will be equal to 0.

The next two values (arccos(-2/3) and -arccos(-2/3) result in the following:

3sin(arccos(-2/3))+2sin(arccos(-2/3))/cos(arccos(-2/3)) =

= sin(arccos(-2/3))*[3+2/(-2/3)] =

=sin(arccos(-2/3))*[3-3] = 0

3sin(-arccos(-2/3))+2sin(-arccos(-2/3))/cos(-arccos(-2/3)) =

= sin(-arccos(-2/3))*[3+2/(-2/3)] =

=sin(-arccos(-2/3))*[3-3] = 0

All is checked.