document.write( "Question 1030354: 3*cos(beta) + 3 = 2*sin^2(beta) \n" ); document.write( "
Algebra.Com's Answer #645236 by ikleyn(52809)\"\" \"About 
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document.write( "\"3%2Acos%28beta%29+%2B+3\" = \"2sin%5E2%28beta%29\".\r\n" );
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document.write( "Replace  \"sin%5E2%28beta%29\"  by  \"1+-+cos%5E2%28beta%29\".  You will get\r\n" );
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document.write( "\"3%2Acos%28beta%29+%2B+3\" = \"2%2A%281+-+cos%5E2%28beta%29%29\",   or\r\n" );
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document.write( "\"3%2Acos%28beta%29+%2B+3\" = \"2+-+2%2Acos%5E2%28beta%29\",   or\r\n" );
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document.write( "\"2%2Acos%5E2%28beta%29+%2B+3%2Acos%28beta%29+%2B+1\" = \"0\",   or  (after factoring left side)\r\n" );
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document.write( "\"%282%2Acos%28beta%29+%2B+1%29%2A%28cos%28beta%29+%2B+1%29\" = \"0\".\r\n" );
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document.write( "It gives you two equations.\r\n" );
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document.write( "1)  \"2%2Acos%28beta%29+%2B+1\" = \"0\"  --->  \"cos%28beta%29\" = \"-1%2F2\"  --->  beta = \"2pi%2F3+%2B+2k%2Api\"  or  beta = \"4pi%2F3+%2B+2k%2Api\",  k = 0, +/-1, +/-2, . . . \r\n" );
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document.write( "2)  \"cos%28beta%29+%2B+1\" = \"0\"  --->  \"cos%28beta%29\" = \"-1\"  --->  beta = \"pi+%2B+2k%2Api\" = \"%282k%2B1%29%2Api\",  k = 0, +/-1, +/-2, . . . \r\n" );
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