The value of θ, lying between θ=0 and θ=π2 and satisfying the equation 1+cos2⁡θsin2⁡θ4sin⁡4θcos2⁡θ1+sin2⁡θ4sin⁡4θcos2⁡θsin2⁡θ1+4sin⁡4θ=0, is

The given equation can be written as1+cos2⁡θsin2⁡θ4sin⁡4θ−1100−11=0 [Applying R3→R3−R2 and R2→R2−R1 ] ⇒ 2sin2⁡θ4sin⁡4θ010−1−11=0 [Applying C1→C1+C2 ] ⇒ 2+4sin⁡4θ=0⇒sin⁡4θ=−12⇒ 4θ=ππ+(−1)n−π6⇒θ=nπ4+(−1)n+1π24We have to choose values of θ s.t 0<θ<π2∴ θ=7π24,11π24

The value of θ, lying between θ=0 and θ=π2 and satisfying the equation 1+cos2⁡θsin2⁡θ4sin⁡4θcos2⁡θ1+sin2⁡θ4sin⁡4θcos2⁡θsin2⁡θ1+4sin⁡4θ=0, is

The given equation can be written as1+cos2⁡θsin2⁡θ4sin⁡4θ−1100−11=0 [Applying R3→R3−R2 and R2→R2−R1 ] ⇒ 2sin2⁡θ4sin⁡4θ010−1−11=0 [Applying C1→C1+C2 ] ⇒ 2+4sin⁡4θ=0⇒sin⁡4θ=−12⇒ 4θ=ππ+(−1)n−π6⇒θ=nπ4+(−1)n+1π24We have to choose values of θ s.t 0<θ<π2∴ θ=7π24,11π24

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The value of $θ$ , lying between $θ$ =0 and $θ = \frac{π}{2}$ and satisfying the equation $|\begin{array}{ccc} 1 + \cos^{2} θ & \sin^{2} θ & 4 \sin 4 θ \\ \cos^{2} θ & 1 + \sin^{2} θ & 4 \sin 4 θ \\ \cos^{2} θ & \sin^{2} θ & 1 + 4 \sin 4 θ \end{array}| = 0$ , is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{11 π}{24}$

$\frac{7 π}{24}$

$\frac{5 π}{24}$

None of these

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The given equation can be written as
$|\begin{array}{ccc} 1 + \cos^{2} θ & \sin^{2} θ & 4 \sin 4 θ \\ - 1 & 1 & 0 \\ 0 & - 1 & 1 \end{array}| = 0$ $[Applying R_{3} \to R_{3} - R_{2} and R_{2} \to R_{2} - R_{1}]$
$\Rightarrow |\begin{array}{ccc} 2 & \sin^{2} θ & 4 \sin 4 θ \\ 0 & 1 & 0 \\ - 1 & - 1 & 1 \end{array}| = 0 [Applying C_{1} \to C_{1} + C_{2}]$
$\begin{array}{l} \Rightarrow 2 + 4 \sin 4 θ = 0 \Rightarrow \sin 4 θ = - \frac{1}{2} \\ \Rightarrow 4 θ = ππ + (- 1)^{n} (- \frac{π}{6}) \Rightarrow θ = \frac{nπ}{4} + (- 1)^{n + 1} \frac{π}{24} \end{array}$
We have to choose values of $θ s.t 0 < θ < \frac{π}{2}$
$∴ θ = \frac{7 π}{24}, \frac{11 π}{24}$