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A value of θ lying between θ=0 and θ=π/2 and satisfying the equation1+sin2θcos2θ4sin4θsin2θ1+cos2θ4sin4θsin2θcos2θ1+4sin4θ=0 is

 

a
3π/24
b
5π/24
c
11π/24
d
π/24

detailed solution

Correct option is C

Applying R1→R1−R3,R2→R2−R3 to the given determinant we get10−101−1sin2⁡θcos2⁡θ1+4sin⁡4θ=0⇒1+4sin⁡4θ+cos2⁡θ+sin2⁡θ=0⇒sin⁡4θ=−1/2⇒4θ=π+π/6 or 2π−π/6 [∵0<4θ<2π]⇒ θ=7π/24 or  11π/24

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