document.write( "Question 410944: solve :(sin B)(sinB)-3sin B+2=0 where 0 < B < 2pie . \n" ); document.write( "

Algebra.Com's Answer #288886 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let $\"z+=+sin%28B%29\"$ . Then, we have a quadratic\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"z%5E2+-+3z+%2B+2+=+0\"$ . Factoring, this is equal to\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " $\"%28z-2%29%28z-1%29+=+0\"$ --> z = 2 or z = 1, sin(B) = 2 or sin(B) = 1 by back-substituting. However, sin(B) cannot be 2 since it is bounded by $\"-1+%3C=+sin%28B%29+%3C=+1\"$ . Therefore, the only solutions occur when $\"sin%28B%29+=+1\"$ --> $\"B+=+pi%2F2\"$ . \n" ); document.write( "

\n" );