document.write( "Question 1172108: Find the values of Arcsin 2x in the equation :
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Algebra.Com's Answer #797066 by ikleyn(52810)\"\" \"About 
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document.write( "Let  \"a\"  be the angle  - \"pi%2F2\" <= a <= \"pi%2F2\",  sin(a) = x,  and \r\n" );
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document.write( "Let  \"b\"  be the angle  - \"pi%2F2\" <= b <= \"pi%2F2\",  sin(b) = 2x.\r\n" );
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document.write( "They want you find the value of x such that\r\n" );
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document.write( "    b - a = \"pi%2F3\".      (1)\r\n" );
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document.write( "Since sin(a) = x  and  sin(b) = 2x,  we have  cos(a) = \"sqrt%281-x%5E2%29\",  cos(b) = \"sqrt%281-4x%5E2%29\".\r\n" );
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document.write( "From equation (1), taking cosine from both sides, we have this equation\r\n" );
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document.write( "    cos(b-a) = \"cos%28pi%2F3%29\",                  or\r\n" );
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document.write( "    cos(b)*cos(a) + sin(b)*sin(a) = \"1%2F2\",   or, substituting\r\n" );
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document.write( "    \"sqrt%281-4x%5E2%29%2Asqrt%281-x%5E2%29\" + (2x)*x = \"1%2F2\",     or\r\n" );
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document.write( "    \"sqrt%28%281-4x%5E2%29%2A%281-x%5E2%29%29\" = \"1%2F2\" - 2x^2.\r\n" );
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document.write( "Now square both sides.  You will get then\r\n" );
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document.write( "    (1-4x^2)*(1-x^2) = \"1%2F4\" - 2x^2 + 4x^4.\r\n" );
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document.write( "Simplify it step by step\r\n" );
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document.write( "    1 - 4x^2 - x^2 + 4x^4 = \"1%2F4\" - 2x^2 + 4x^4\r\n" );
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document.write( "    1 - 5x^2              = \"1%2F4\" - 2x^2\r\n" );
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document.write( "    4 - 20x^2             = 1 - 8x^2\r\n" );
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document.write( "    4 - 1                 = 20x^2 - 8x^2\r\n" );
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document.write( "      3                   = 12x^2\r\n" );
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document.write( "      1                   = 4x^2\r\n" );
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document.write( "      x^2                 = \"1%2F4\"\r\n" );
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document.write( "      x                   = \"sqrt%281%2F4%29%29\" = +/- \"1%2F2\".\r\n" );
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document.write( "Thus the equation is just solved,  and we have two potential solutions  x= +/- \"1%2F2\".\r\n" );
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document.write( "    Consider these two cases separately and check the results in both cases.\r\n" );
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document.write( "Case a).  x = \"1%2F2\";  arcsin(x) = \"pi%2F6\";  2x = 1;  arcsin(2x) = arcsin(1) = \"pi%2F2\".\r\n" );
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document.write( "          Since  \"pi%2F2\" - \"pi%2F6\" = \"3pi%2F6+-+pi%2F6\" = \"2pi%2F6\" = \"pi%2F3\",  the solution is correct.\r\n" );
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document.write( "Case b).  x = - \"1%2F2\";  arcsin(x) = - \"pi%2F6\";  2x = -1;  arcsin(2x) = arcsin(-1) = - \"pi%2F2\".\r\n" );
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document.write( "          Since  \"-pi%2F2\" - \"-pi%2F6\" = \"-3pi%2F6+%2B+pi%2F6\" = - \"2pi%2F6\" = - \"pi%2F3\",  this solution DOES NOT work. It is EXTRANEOUS.\r\n" );
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document.write( "   +--------------------------------------------------------------------+\r\n" );
document.write( "   |  So, the problem has a unique solution                             |\r\n" );
document.write( "   |                                                                    |\r\n" );
document.write( "   |  x = \"1%2F2\",  and  arcsin(x) = \"pi%2F6\",  arcsin(2x) = \"pi%2F2\".                 | \r\n" );
document.write( "   +--------------------------------------------------------------------+\r\n" );
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\n" ); document.write( "\n" ); document.write( "Post-solution notes:\r
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document.write( "1)  The solution under the link\r\n" );
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document.write( "    https://socratic.org/questions/how-do-you-solve-arcsin-x-arcsin-2x-pi-3\r\n" );
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document.write( "    mentioned by tutor  @Math_tutor2020, is INCORRECT.\r\n" );
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document.write( "2)  The solution by @Math_tutor2020 also contains a technical error.\r\n" );
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document.write( "    It is WHY I came to bring the correct and accurate solution.\r\n" );
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