Question

A value of `theta in (0, pi//3),` for which
`|{:(1+"cos"^(2)theta, "sin"^(2)theta, 4"cos" 6 theta), ("cos"^(2)theta, 1+"sin"^(2)theta, 4"cos"6theta), ("cos"^(2)theta, "sin"^(2)theta, 1+4"cos"6theta):}| = 0,` is
A. `(pi)/(9)`
B. `(pi)/(18)`
C. `(7pi)/(24)`
D. `(7pi)/(36)`

Answer 1

Correct Answer - A
Let `Delta = |{:(1+"cos"^(2)theta, "sin"^(2)theta, 4"cos"6theta),("cos"^(2)theta, 1+"sin"^(2)theta, 4"cos"6theta), ("cos"^(2)theta, "sin"^(2)theta, 1+4"cos"6theta):}| =0`
`"Applying "C_(1) to C_(1) + C_(2),` we get
`Delta = |{:(2, "sin"^(2)theta, 4"cos"6theta),(2, 1+"sin"^(2)theta, 4"cos"6theta), (1, "sin"^(2)theta, 1+4"cos"6theta):}| =0`
Applying `R_(1) to R_(1) - 2R_(3) " and " R_(2) to R_(2) - 2R_(3),` we get
`Delta = |{:(0, -"sin"^(2)theta, -2-4"cos"6theta),(0, 1-"sin"^(2)theta, -2-4"cos"6theta), (1, "sin"^(2)theta, 1+4"cos"6theta):}| =0`
On expanding w.r.c `C_(1)`, we get
`rArr "sin"^(2)theta (2 + 4 "cos"6theta) + (2 + 4 "cos"6theta) (1-"sin"^(2)theta) = 0`
`rArr 2 +4 "cos" 6 theta = 0 rArr "cos"6theta = -(1)/(2) = "cos" (2x)/(3)`
`rArr 6 theta = (2x)/(3) rArr theta = (pi)/(9) " "[because theta in (0, (pi)/(3))]`