document.write( "Question 1036084: For questions 1-2 solve each equation on the interval [0,2pi). Show all work.\r
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document.write( "1. 4 sin^2 x + 2 sin x- 2 = 0\r
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document.write( "2. 2 sin(3x) = 1 \n" );
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Algebra.Com's Answer #650698 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! 4 sin^2 x + 2 sin x- 2 = 0 \n" ); document.write( "factor out a 2. \n" ); document.write( "2(2sin^2 x+ sin x -1)=0 \n" ); document.write( "(2sinx-1)(sin x +1)=0 \n" ); document.write( "2 sin x -1=0, 2 sin x=1, sin x =(1/2), where it is at pi/6 and 5 pi/6. \n" ); document.write( "sin x+1=0 \n" ); document.write( "sin x=-1 \n" ); document.write( "That is the case at 3pi/2. \n" ); document.write( "==================== \n" ); document.write( "2sin (3x)=1 \n" ); document.write( "sin (3x)=(1/2) \n" ); document.write( "The solutions above are divided by 3, so it occurs at 3x=pi/6, and pi/18 and 5pi/18 would be the solutions.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |