document.write( "Question 1036275: sin(3x)+cos(3x) = -1 solve for x. \n" ); document.write( "

Algebra.Com's Answer #650982 by ikleyn(52787) $\"\"$ $\"About$

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document.write( "sin(3x) + cos(3x) = -1.    (1)   (It is the original equation)\r\n" );
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document.write( "Square its both sides. You will get\r\n" );
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document.write( " = .   (2)\r\n" );
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document.write( "From the other side, there is an identity\r\n" );
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document.write( " == .   (3)\r\n" );
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document.write( "Comparing (2) and (3), you get\r\n" );
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document.write( "2*sin(3x)*cos(3x) = 0,   or  sin(3x)*cos(3x) = 0.    (4)\r\n" );
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document.write( "Equation (4) splits in two independent equations\r\n" );
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document.write( "1)  sin(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . \r\n" );
document.write( "                 --->   x = ,  k = 0, +/-1. +/-2, . . .       (5)\r\n" );
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document.write( "2)  cos(3x) = 0  --->  3x = ,   k = 0. +/-1. +/-2, . . . \r\n" );
document.write( "                 --->   x = ,  k = 0. +/-1. +/-2, . . .   (6)\r\n" );
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document.write( "Now we should check which of the found values (5), (6) satisfy the original equation.\r\n" );
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document.write( "Of the set (5), all x satisfy sin(3x) = 0. Hence, only those of (5) satisfy the original equation where cos(3x) = -1.\r\n" );
document.write( "They are  3x = ,    n = 0, +/-1. +/-2, . . . ,  or \r\n" );
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document.write( "           x = ,  n = 0, +/-1. +/-2, . . . ,           (5').\r\n" );
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document.write( "Of the set (6), all x satisfy cos(3x) = 0. Hence, only those of (6) satisfy the original equation where sin(3x) = -1.\r\n" );
document.write( "They are  3x = ,   n = 0, +/-1. +/-2, . . . ,  or \r\n" );
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document.write( "           x = ,  n = 0, +/-1. +/-2, . . . ,           (6'). \r\n" );
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document.write( "Answer.  The union of the sets (5') and (6') is the solution of the original equation. \r\n" );
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