document.write( "Question 594181: find solutions of sqrt(3)cos(x)+sin(x)=1, on the interval[0,2pi) \n" ); document.write( "
Algebra.Com's Answer #376759 by lwsshak3(11628)\"\" \"About 
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find solutions of sqrt(3)cos(x)+sin(x)=1, on the interval[0,2pi
\n" ); document.write( "√3cos+sin=1
\n" ); document.write( "√3cos=1-sin
\n" ); document.write( "square both sides
\n" ); document.write( "3cos^2=1-2sin+sin^2
\n" ); document.write( "3cos^2=1-2sin+1-cos^2
\n" ); document.write( "4cos^2=2-2sin
\n" ); document.write( "divide by 2
\n" ); document.write( "2cos^2=1-sin
\n" ); document.write( "2(1-sin^2)=1-sin
\n" ); document.write( "2-2sin^2=1-sin
\n" ); document.write( "2sin^2-sin-1=0
\n" ); document.write( "(2sinx+1)(sinx-1)
\n" ); document.write( "..
\n" ); document.write( "2sinx+1=0
\n" ); document.write( "sinx=-1/2
\n" ); document.write( "x=11π/6
\n" ); document.write( "or
\n" ); document.write( "sinx-1=0
\n" ); document.write( "sinx=1
\n" ); document.write( "x=π/2
\n" ); document.write( "..
\n" ); document.write( "Check:
\n" ); document.write( "For x=11π/6
\n" ); document.write( "√3cosx+sinx=√3*√3/2-1/2=3/2-1/2=1
\n" ); document.write( "..
\n" ); document.write( "for x=π/2
\n" ); document.write( "√3cosx+sinx=0+1=1\r
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