Question 1172108
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Find the values of Arcsin 2x in the equation :
Arcsin 2x - Arcsin x = pi/3
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<pre>
Let  "a"  be the angle  - {{{pi/2}}} <= a <= {{{pi/2}}},  sin(a) = x,  and 

Let  "b"  be the angle  - {{{pi/2}}} <= b <= {{{pi/2}}},  sin(b) = 2x.


They want you find the value of x such that

    b - a = {{{pi/3}}}.      (1)


Since sin(a) = x  and  sin(b) = 2x,  we have  cos(a) = {{{sqrt(1-x^2)}}},  cos(b) = {{{sqrt(1-4x^2)}}}.


From equation (1), taking cosine from both sides, we have this equation

    cos(b-a) = {{{cos(pi/3)}}},                  or

    cos(b)*cos(a) + sin(b)*sin(a) = {{{1/2}}},   or, substituting

    {{{sqrt(1-4x^2)*sqrt(1-x^2)}}} + (2x)*x = {{{1/2}}},     or

    {{{sqrt((1-4x^2)*(1-x^2))}}} = {{{1/2}}} - 2x^2.


Now square both sides.  You will get then

    (1-4x^2)*(1-x^2) = {{{1/4}}} - 2x^2 + 4x^4.


Simplify it step by step

    1 - 4x^2 - x^2 + 4x^4 = {{{1/4}}} - 2x^2 + 4x^4

    1 - 5x^2              = {{{1/4}}} - 2x^2

    4 - 20x^2             = 1 - 8x^2

    4 - 1                 = 20x^2 - 8x^2

      3                   = 12x^2

      1                   = 4x^2

      x^2                 = {{{1/4}}}

      x                   = {{{sqrt(1/4))}}} = +/- {{{1/2}}}.


Thus the equation is just solved,  and we have two potential solutions  x= +/- {{{1/2}}}.


    Consider these two cases separately and check the results in both cases.



Case a).  x = {{{1/2}}};  arcsin(x) = {{{pi/6}}};  2x = 1;  arcsin(2x) = arcsin(1) = {{{pi/2}}}.

          Since  {{{pi/2}}} - {{{pi/6}}} = {{{3pi/6 - pi/6}}} = {{{2pi/6}}} = {{{pi/3}}},  the solution is correct.



Case b).  x = - {{{1/2}}};  arcsin(x) = - {{{pi/6}}};  2x = -1;  arcsin(2x) = arcsin(-1) = - {{{pi/2}}}.

          Since  {{{-pi/2}}} - {{{-pi/6}}} = {{{-3pi/6 + pi/6}}} = - {{{2pi/6}}} = - {{{pi/3}}},  this solution DOES NOT work. It is EXTRANEOUS.


   +--------------------------------------------------------------------+
   |  So, the problem has a unique solution                             |
   |                                                                    |
   |  x = {{{1/2}}},  and  arcsin(x) = {{{pi/6}}},  arcsin(2x) = {{{pi/2}}}.                 | 
   +--------------------------------------------------------------------+
</pre>


Solved.



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Post-solution notes:


<pre>
1)  The solution under the link

    <a href = "https://socratic.org/questions/how-do-you-solve-arcsin-x-arcsin-2x-pi-3">https://socratic.org/questions/how-do-you-solve-arcsin-x-arcsin-2x-pi-3</a>

    mentioned by tutor  @Math_tutor2020, is INCORRECT.



2)  The solution by @Math_tutor2020 also contains a technical error.

    It is WHY I came to bring the correct and accurate solution.
</pre>