Question 1176724
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Added after seeing the response from tutor @ikleyn....<br>
She used a nice fact that I overlooked that makes solving this problem relatively easy.<br>
The numerator is the sum of two angles, each less than pi/2 or 90 degrees, in which the sine of one angle is the cosine of the other.  But that makes the two angles complementary, so the numerator of the fraction is just pi/2 (radians), or 90 degrees.<br>
Then from there the solution is relatively easy.<br>
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To solve this algebraically, I think you would have to use calculus and the Taylor series expressions for arcsin, arccos, and arctan.  And even then solving the problem would probably involve very ugly calculations.<br>
On the other hand, solving the problem using a graphing calculator is easy -- simply graph the expressions on the two sides of the equation and find where they are equal.<br>
Note that the domains of arcsin and arctan are from -pi/2 to pi/2 and the domain of arccos is from 0 to pi; so the domain of the expression shown is from 0 to pi/2.<br>
HOWEVER....<br>
The topic you have chosen for your post is trigonometry basics -- and solving the problem either of those ways is not basics.<br>
If solving the problem involves basic trigonometry, then the answer likely lies with one of the "nice" acute angles.  And for "nice" acute angles, with the "x/2" as the argument for the sine and cosine and "x" as the argument for tangent, x=sqrt(3) seems a likely candidate.<br>
And, indeed, with the angles in degrees,<br>
{{{arcsin(sqrt(3)/2) = 60}}}
{{{arccos(sqrt(3)/2) = 30}}}
{{{arctan(sqrt(3)) = 60}}}<br>
and<br>
{{{(arcsin(sqrt(3)/2)+arccos(sqrt(3)/2))/arctan(sqrt(3)) = (60+30)/60 = 3/2}}}<br>
ANSWER: x = sqrt(3)<br>