SOLUTION: Give the general solution to the equation sinx+sin2x+sin3x=0

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Question 1069081: Give the general solution to the equation sinx+sin2x+sin3x=0
Found 2 solutions by t0hierry, ikleyn:
Answer by t0hierry(194) About Me  (Show Source):
You can put this solution on YOUR website!
sin x + sin 2x + sing 3x = ((cos x/2 - cos(3 + 1/2)x)/2sin(x/2) = 0
-2sin(2x)sin(3x/2)/2sin(x/2) = 0
Solutions x = (2n+1)pi odd multiples of pi

Answer by ikleyn(52832) About Me  (Show Source):
You can put this solution on YOUR website!
.
Give the general solution to the equation sinx+sin2x+sin3x=0
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sin(x) + sin(2x) + sin(3x) = 0.     (1)


Apply the trigonometry formula  sin%28a%29+%2B+sin%28b%29 = 2%2Asin%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29

     (see any serious textbook in trigonometry or the lessons 

         - FORMULAS FOR TRIGONOMETRIC FUNCTIONS
         - Addition and subtraction of trigonometric functions

      in this site) to the first and third addend in the left side of the original equation (1).   You will get

sin%28x%29%2Bsin%283x%29 = 2%2Asin%28%28x%2B3x%29%2F2%29%2Acos%28%28x-3x%29%2F2%29 = 2%2Asin%282x%29%2Acos%28-x%29 = 2%2Asin%282x%29%2Acos%28x%29.

Now, the equation (1) takes the form

2%2Asin%282x%29%2Acos%28x%29+%2B+sin%282x%29 = 0,   or

sin%282x%29%2A%282%2Acos%28x%29+%2B1%29 = 0.

This equation deploys in two independent equations


1.  sin(2x) = 0  --->  x = k%2Api, x = pi%2F2+%2B+k%2Api,  k = 0, =/-1, +/-2. . . . 


2.  2cos(x) + 1 = 0  --->  cos(x) = -1%2F2  --->  x = 2pi%2F3+%2B+2k%2Api,  x = 4pi%2F3+%2B+2k%2Api,  k = 0, =/-1, +/-2. . . . 


Answer.  The solutions are  a)  x = k%2Api, x = pi%2F2+%2B+k%2Api,  k = 0, =/-1, +/-2. . . .    and 

                            b)  x = 2pi%2F3+%2B+2k%2Api,  x = 4pi%2F3+%2B+2k%2Api,  k = 0, =/-1, +/-2. . . . 





Plot y = sin(x) + sin(2x) + sin(3x)

The solution by the other tutor is incorrect.