Solving Trigonometric Equations

Category: Trigonometry, Algebra

"Published in Newark, California, USA"

Find the values of x in the range from 0º to 360º for

                   Sin x + Sin 2x + Sin 3x = 0

Solution:

Consider the given equation

                    Sin x + Sin 2x + Sin 3x = 0

Apply the Sum and Difference of Two Angles Formula and Double Angle Formula for the above equation

                  Sin x + 2 Sin x Cos x + Sin (x + 2x) = 0 

         Sin x + 2 Sin x Cos x + Sin x Cos 2x + Cos x Sin 2x = 0

 Sin x + 2 Sin x Cos x + Sin x (Cos² x - Sin² x) + Cos x (2 Sin x Cos x) = 0

       Sin x + 2 Sin x Cos x + Sin x Cos² x - Sin³ x + 2 Sin x Cos² x = 0

                Sin x + 2 Sin x Cos x + 3 Sin x Cos² x - Sin³ x = 0

Take out their common factor which is Sin x, we have

               Sin x (1 + 2 Cos x + 3 Cos² x - Sin² x) = 0

               Sin x [1 + 2 Cos x + 3 Cos² x - (1 - Cos² x)] = 0

               Sin x (1 + 2 Cos x + 3 Cos² x - 1 + Cos² x) = 0

               Sin x (2 Cos x + 4 Cos² x) = 0

               (Sin x )(2 Cos x)(1 + 2 Cos x) = 0

Equate each factor in zero

For          Sin x = 0

               x = Sin^-1 0
     
               x = 0º, 180º, 360º

For          2 Cos x = 0

               Cos x = 0

               x = Cos^-1 0

               x = 90º, 270º

For         1 + 2 Cos x = 0

               2 Cos x = -1

               Cos x = - ½

               x = Cos^-1 -½

               x = 120º, 240º

Therefore, 

               x = 0º, 90º, 120º, 180º, 240º, 270º, and 360º