Three consecutive integers have a sum of 234 what are the three integers
1 answer:
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<h2><u>Answer:</u></h2>
x=(nπ/2)±15
x=(2nπ+90)/3
x=90-2nπ
<h2><u>Steps:</u></h2>cos3x+sin2x-sin6x+cos5x=0
(cos3x+cos5x)+(sin2x-sin6x)=0
equation(1)
<u>Use formula:</u>
<u>So in equation (1),if c=3x ,d=5x,C=2x,D=6x</u>
》2cos{(c+d)/2}.cos{(c-d)/2}+2cos{(C+D)/2}.sin{(C-D)/2}=0
》2cos{(3x+5x)/2}.cos{(3x-5x)/2}+2cos{(2x+6x)/2}.sin{(2x-6x)/2}=0
》2cos(4x).cos(-x)+2cos(4x).sin(-2x)=0
》2cos(4x)[cos(-x)+sin(-2x)]=0
》2cos(4x)[cos(x)-sin(2x)]=0
<u>1)</u><u> </u><u>Either:</u>
2cos(4x)=0
cos(4x)=0/2
cos(4x)=0
cos(4x)=cos(90)
General solution for such case is:X=2n±a
So,
4x=2nπ±90
x=(2nπ±90)/4
x=(nπ/2)±15
<u>2) Or:</u>
cos(x)-sin(2x)=0
cos(x)=sin(2x)
General solution for such case is:X=2n±a
So,
x=2nπ±(90-2x)
x=2nπ±90±2x
x±2x=2nπ±90
<u>Take +ve sign,</u>
x+2x=2nπ+90
3x=2nπ+90
x=(2nπ+90)/3
<u>Take -ve sign,</u>
x-2x=2nπ-90
-x=2nπ-90
x=(2nπ-90)/(-1)
x=90-2nπ