Question 1176724
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if x > 0,      solve for x      {{{(arcsin(x/2) + arccos(x/2))/arctan(x)}}} = {{{3/2}}}.


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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Solution</U>



<pre>
Let a = {{{arcsin(x/2)}}},  b = {{{arccos(x/2)}}}.


It means that  


    sin(a) = {{{x/2}}};  cos(b) = {{{x/2}}}  and  "a"  and  "b"  are the angles in QI  (acute angles).


Since  sin(a) = cos(b),  it means that  a + b = {{{pi/2}}}.


In other words, the numerator in our fraction, which is  {{{arcsin(x/2)}}} + {{{arccos(x/2)}}}, is equal to  {{{pi/2}}}.


Thus our original equation is  {{{((pi/2))/arctan(x)}}} = {{{3/2}}}.


It implies  arctan(x) = {{{((pi/2))/((3/2))}}} = {{{pi/3}}}.


Since  arctan(x) = {{{pi/3}}},  it implies  x = {{{tan(pi/3)}}} = {{{sqrt(3)}}}.


<U>ANSWER</U>.  x = {{{sqrt(3)}}}.
</pre>

Solved.