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Question

Consider the system of equations
xcos3y+3xcosysin2y=14
xsin3y+3xcos2ysiny=13

The value of sin2y+2cos2y is

A
45
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B
95
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C
2
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D
None of these
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Solution

The correct option is B 95
xcos3y+3xcosysin2y=14 (1)
xsin3y+3xcos2ysiny=13 (2)
Adding Eqs. (1) and (2)
x(cos3y+3cos2ysiny+3cosysin2y+sin3y)=27
x(cosy+siny)3=27
x1/3(cosy+siny)=3 (3)

Subtracting Eq (2) from Eq (1), we get
x(cos3y+3cosysin2y3cos2ysinysin3y)=1
x(cosysiny)3=1
x1/3(cosysiny)=1 (4)

Dividing Eq (3) by Eq (4), we get
cosy+siny=3cosy3siny
tany=12
siny=15,cosy=25
or siny=15,cosy=25

sin2y+2cos2y =15+85=95


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