Question 1029560
.
√3*sin(x/2)+cos(x/2)=0

(&#960;&#8804;x<4&#960;)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{sqrt(3)*sin(x/2) + cos(x/2)}}} = {{{0}}}  --->  (divide bith sides by 2)  ---> 

{{{(sqrt(3)/2)*sin(x/2) + (1/2)*cos(x/2)}}} = {{{0}}}.   (1)

Notice that {{{sqrt(3)/2}}} = {{{cos(pi/6)}}}  and  {{{1/2}}} = {{{sin(pi/6)}}}.

Therefore, you can rewrite (1) as

{{{cos(pi/6)*sin(x/2) + sin(pi/6)*cos(x/2)}}} = {{{0}}}.   (2)

Now apply the addition formula for sine {{{cos(alpha)*sin(beta)+sin(alpha)*cos(beta)}}} = {{{sin(alpha + beta)}}}.

You will get instead of (2)

{{{sin(pi/6 + x/2)}}} = {{{0}}}.

Then  pi/6 + x/2 = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 

Hence x = {{{2k*pi - pi/3}}}.

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

You will have two solutions  x = {{{2pi - pi/3}}} = {{{5pi/3}}}  and  x = {{{4pi - pi/3}}} = {{{11pi/3}}}.

<U>Answer</U>. There are two solutions in the given interval. 

        They are  x = {{{5pi/3}}}  and x = {{{11pi/3}}}.
</pre>