document.write( "Question 374824: how to solve 2sin^2x-3sinx+1=0 \n" ); document.write( "
Algebra.Com's Answer #266550 by vasumathi(46)\"\" \"About 
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solve 2sin^2x-3sinx+1=0
\n" ); document.write( "Solution : Let sinx = u
\n" ); document.write( "plug in this in the given equation
\n" ); document.write( "2u^2-3u+1 = 0
\n" ); document.write( "This canbe factored as follows...
\n" ); document.write( "(2u-1) ( u-1)= 0
\n" ); document.write( "2u-1 = 0 -> 2sinx - 1 = 0-----> sinx = 1/2---> x = pie/6 Since this is periodic with a period 2 pie we have (2*n*pie+pie/6) as a complete solution
\n" ); document.write( "now ...
\n" ); document.write( "u-1 = 0 --->sinx-1=0-> sinx = 1-------->x = pie/2 .Since this is periodic with a period 2 pie we have (2*n*pie+pie/2) as a complete solution
\n" ); document.write( "So the solutions are 1)(2*n*pie+pie/6)
\n" ); document.write( " 2)(2*n*pie+pie/2)\r
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