Consider the system of equations
`x cos^(3) y+3x cos y sin^(2) y=14`
`x sin^(3) y+3x cos^(2) y sin y=13`
The value/values of x is/are

Options

A. `4//5`
B. `9//5`
C. 2
D. none of these

Answer

B

Explanation

The given equations are
`x cos^(3)y+3x cos y sin^(2) y=14` ...(i)
and `x sin^(2) y+3x cos^(2) y sin y=13` ...(ii)
Adding Eqs. (i) and (ii), we have
`x(cos^(3) y+3 cos y sin^(2) y+3 cos^(2) y sin y+ sin^(3) y)=27`
or `x(cos y+ sin y)^(3)=27`
or `x^(1//3) (cos y + sin y) =3` ...(iii)
Subtracting Eq. (ii) from Eq. (i), we have
`x(cos^(3)y+3 cos y sin^(2) y-3 cos^(2) y sin y- sin^(3) y)=1`
or `x(cos y- sin y)^(3)=1`
or `x^(1//3) (cos y- sin y)=1` ...(iv)
Dividing Eq. (iii) by (iv), we get
`cos y+sin y=3 cos y-3 sin y`
or `tan y=1//2`
Case I :
`sin y=1//sqrt(5) and cos y =2//sqrt(5)`
`y=2n pi +alpha`, where `0 lt alpha lt pi//2` and `sin alpha =1//sqrt(5)`
i.e., y lies in the first quadrant
From Eqs. (iii) `x^(1//3) (3//sqrt(5))=3 or x=5 sqrt(5)`
Case II :
`sin y=-1//sqrt(5) and cos y=-2//sqrt(5)`
`y=2npi+(pi+alpha)`, where `0 lt alpha lt pi//2`
and `sin alpha = -1 //sqrt(5)`
i.e., y lies in the third quadrant.
Therefore, from Eq. (iii), `x^(1//3) (-3//sqrt(5))=3 or x=-5sqrt(5)`.
Thus, `sin^(2) y+2 cos^(@) y=1//5+8//5=9//5`.
Also there are exactly six values of `y in [0, 6pi]`, there in 1st quadrant and three in 3rd quadrant.