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Question

Solve:
∣ ∣ ∣1+sin2θcos2θ4sin4θsin2θ1+cos2θ4sin4θsin2θcos2θ1+4sin4θ∣ ∣ ∣ =0; where, 0<θ<π2.

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Solution

Given ∣ ∣ ∣1+sin2θcos2θ4sin4θsin2θ1+cos2θ4sin4θsin2θcos2θ1+4sin4θ∣ ∣ ∣ =0; where, 0<θ<π2.
(C1C1+C2)

∣ ∣ ∣2cos2θ4sin4θ21+cos2θ4sin4θ1cos2θ1+4sin4θ∣ ∣ ∣ =0

(R2R2R1 & R3R3R1)

∣ ∣2cos2θ4sin4θ010101∣ ∣ =0
2+4sin4θ=0

sin4θ=12

4θ=7π6 ...[Since 0<θ<π2]
θ=7π24

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