document.write( "Question 731700: Hi,
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document.write( "Can someone tell me the steps how to solve this\r
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document.write( "cox x + 1 = √3 sin x where 0 ≤ x < 2∏\r
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document.write( "the answer is cos x = 1/2 or cos x =-1\r
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document.write( "I just want to know how did they get it.\r
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document.write( "Thank you so much in advance. \n" );
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Algebra.Com's Answer #447245 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! cos x + 1 = √3 sin x where 0 ≤ x < 2π \n" ); document.write( "cos x + 1 = √3 sin x \n" ); document.write( "square both sides \n" ); document.write( "cos^2x+2cosx+1=3sin^2x \n" ); document.write( "cos^2x+2cosx+1=3(1-cos^2x) \n" ); document.write( "cos^2x+2cosx+1=3-3cos^2x) \n" ); document.write( "4cos^2x+2cosx-2=0 \n" ); document.write( "(4cosx-2)(cosx+1)=0 \n" ); document.write( ".. \n" ); document.write( "4cosx-2=0 \n" ); document.write( "cosx=1/2 \n" ); document.write( "x=π/3, 5π/3 (In quadrants I and IV where cos>0) \n" ); document.write( "or \n" ); document.write( "cosx+1=0 \n" ); document.write( "cosx=-1 \n" ); document.write( "x=π \n" ); document.write( " |