SOLUTION: solutions of sin5x + sin3x=0 in the interval [0, 2\pi ]

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Question 1105800: solutions of sin5x + sin3x=0 in the interval [0, 2\pi ]
Answer by ikleyn(52878)   (Show Source): You can put this solution on YOUR website!
.
Use the formula  sin(a) + sin(b) = 2*sin((a+b)/2)*cos((a-b)/2),

which is one of basic formula of Trigonometry.



In your case  a = 5x,  b = 3x.   Hence,

sin(5x) + sin(3x) =  = 2*sin(4x)*cos(x).



Therefore, the original equation is equivalent to

2(sin(4x)*cos(x) = 0.



The last equation deploys in two equations:


    1)  sin(4x) = 0,  which in the given interval has the solutions

        x= 0,  ,    and  ,

and

    2)  cos(x) = 0,  which in the given interval has the solutions

        x=   and  .



Answer.  The four solutions to the original equation are x= 0,  ,    and  .


------------
On relevant formulas of trigonometry see the lessons
    - FORMULAS FOR TRIGONOMETRIC FUNCTIONS
    - Addition and subtraction of trigonometric functions
    - Addition and subtraction of trigonometric functions - Examples
in this site.


Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Trigonometry. Formulas for trigonometric functions".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.



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