SOLUTION: Solve, for degree between zero and 360, the equation {{{4sinX+3cosX=0}}}

Algebra ->  Trigonometry-basics -> SOLUTION: Solve, for degree between zero and 360, the equation {{{4sinX+3cosX=0}}}      Log On


   

Question 195263: Solve, for degree between zero and 360, the equation 4sinX%2B3cosX=0
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Solve, for degree between zero and 360, the equation
4sin%28X%29%2B3cos%28X%29=0

Divide through by cos%28X%29

4sin%28X%29%2Fcos%28X%29%2B3cos%28X%29%2Fcos%28X%29=0%2Fcos%28X%29

4sin%28X%29%2Fcos%28X%29%2B3=0

Use the identity sin%28alpha%29%2Fcos%28alpha%29=tan%28alpha%29
to replace the left side:

4tan%28X%29%2B3=0

Add -3 to both sides

4tan%28X%29+=+-3

Divide both sides by 4

tan%28X%29+=+-3%2F4

There are two quadrants in which
the tangent is negative, QII and QIV

They are these two angles:

 and this  

Now we find the reference angle of these two
angles by calculator using inverse tangent
and POSITIVE 3%2F4

Tan%5E%28-1%29%283%2F4%29+=+36.86989765°

The angle in the second quadrant is found
by subtracting the reference angle from 180°

180° - 36.86989765° = 143.1301025°

The angle in the fourth quadrant is found
by subtracting the reference angle from 360°

360° - 36.86989765° = 323.1301025°

Edwin