SOLUTION: Find the general solution to sinx+sin2x+sin3x+sin4x=0

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Question 1138303: Find the general solution to sinx+sin2x+sin3x+sin4x=0
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
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sin(x) + sin(2x) + sin(3x) + sin(4x) = 0.        (1)


Use the general formula of Trigonometry

sin%28a%29+%2B+sin%28b%29 = 2sin%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29.                (2)


You have 

sin(x) + sin(4x) = 2%2Asin%28%28x%2B4x%29%2F2%29%2Acos%28%284x-x%29%2F2%29 = 2%2Asin%282.5x%29%2Acos%281.5x%29,

sin(2x) + sin(3x) = 2%2Asin%28%282x%2B3x%29%2F2%29%2Acos%28%283x-2x%29%2F2%29 = 2%2Asin%282.5x%29%2Acos%280.5x%29.


Therefore, the left side of the original equation is

sin(x) + sin(2x) + sin(3x) + sin(4x) = 2*sin(2.5x)*cos(1.5x) + 2*sin(2.5x)*cos(0.5x) = 2*sin(2.5x)*(cos(1.5x) + cos(0.5x)).


Hence, the original equation is equivalent to

2*sin(2.5x)*(cos(1.5x) + cos(0.5x)) = 0,    or,  canceling  the factor  2*sin(2.5x),

cos(1.5x) + cos(0.5x) = 0.                       (3)


Next, apply another general formula of Trigonometry

cos%28a%29+%2B+cos%28b%29 = 2cos%28%28a%2Bb%29%2F2%29%2Acos%28%28a-b%29%2F2%29.                (4)


Then the equation (3) becomes

2%2Acos%28x%29%2Acos%28x%2F2%29 = 0.                                (5)


Equation (5) deploys in two independent separate equations:


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


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

                        x = pi+%2B+2k%2Api = %282k%2B1%29%2Api,  k = 0, +/-1, +/-2, . . . 


From (1) and (2), in the given interval the original equation has the roots pi%2F2, pi, 3pi%2F2,  or  90°,  180°,  270°.


But these are not ALL the roots.

There is one more family of roots.

Do you remember I canceled the factor 2*sin(2.5x) ?

Of course, I must consider (and add to the solution set !) all the solutions of the equation

sin(2.5x) = 0.

They are  2.5x = k%2Api,  k = 0, +/-1, +/-2, . . . 

or, which is the same,

%285x%29%2F2 = k%2Api,  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = 0, %282k%2Api%29%2F5, %284k%2Api%29%2F5, %286k%2Api%29%2F5, %288k%2Api%29%2F5,  k = 0, +/-1, +/-2, . . . 

The final answer is:  There are two families of solutions. 

                      One family is  pi%2F2+%2B+2k%2Api,  pi%2B2k%2Api,  and  3pi%2F2%2B+2k%2Api,  k = 0, +/-1, +/-2, . . . , or  90°, 180° and 270°.

                      The other family is  %282k%2Api%29%2F5,  k = 0, +/-1, +/-2, . . . ,  or  0°, 72°, 144°, 216°, 288°.

Solved.

CHECK

See the plot of the left side of the original equation 




Plot y = sin(x) + sin(2x) + sin(3x) + sin(4x)



Do you see 8 roots in the interval [0,2pi) ?