SOLUTION: &#8730;3*sin(x/2)+cos(x/2)=0 (&#960;&#8804;x<4&#960;)

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Question 1029560: √3*sin(x/2)+cos(x/2)=0
(π≤x<4π)
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.
√3*sin(x/2)+cos(x/2)=0
(π≤x<4π)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

sqrt%283%29%2Asin%28x%2F2%29+%2B+cos%28x%2F2%29 = 0  --->  (divide bith sides by 2)  ---> 

%28sqrt%283%29%2F2%29%2Asin%28x%2F2%29+%2B+%281%2F2%29%2Acos%28x%2F2%29 = 0.   (1)

Notice that sqrt%283%29%2F2 = cos%28pi%2F6%29  and  1%2F2 = sin%28pi%2F6%29.

Therefore, you can rewrite (1) as

cos%28pi%2F6%29%2Asin%28x%2F2%29+%2B+sin%28pi%2F6%29%2Acos%28x%2F2%29 = 0.   (2)

Now apply the addition formula for sine cos%28alpha%29%2Asin%28beta%29%2Bsin%28alpha%29%2Acos%28beta%29 = sin%28alpha+%2B+beta%29.

You will get instead of (2)

sin%28pi%2F6+%2B+x%2F2%29 = 0.

Then  pi/6 + x/2 = k%2Api,  k = 0, +/-1, +/-2, . . . 

Hence x = 2k%2Api+-+pi%2F3.

To get the given interval for x, take k = 1 and k = 2.

You will have two solutions  x = 2pi+-+pi%2F3 = 5pi%2F3  and  x = 4pi+-+pi%2F3 = 11pi%2F3.

Answer. There are two solutions in the given interval. 

        They are  x = 5pi%2F3  and x = 11pi%2F3.