SOLUTION: how do i solve 2sin(x) + sqr3 = 0 for 0 < x < 2pi

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Question 1094229: how do i solve 2sin(x) + sqr3 = 0 for 0 < x < 2pi
Answer by ikleyn(52798) About Me  (Show Source):
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how do i solve 2sin(x) + sqr3 = 0 for 0 < x < 2pi
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2sin(x) + sqrt(3) = 0  ====>

2*sin(x) = -sqrt%283%29  ====>

sin(x) = -sqrt%283%29%2F2  ====>

x = 4pi%2F3   or/and  x = 5pi%2F3.


Answer.  There are TWO solutions:  4pi%2F3  and  5pi%2F3.


For similar or close problems see the lessons
    - Solving simple problems on trigonometric equations
    - Solving typical problems on trigonometric equations
    - Solving more complicated problems on trigonometric equations
    - Solving advanced problems on trigonometric equations
    - Proving Trigonometry identities
    - OVERVIEW of lessons on calculating trig functions and solving trig equations
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: Solved problems".


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.