If 4( sin 2x sin 4x + sin2 x) = 3 then x =

Question

If 4( sin 2x sin 4x + sin2 x) = 3 then x = (A) (2 n π/3) ± (π/9) , n ∈ Z  (B) (n π/3) ± (π/9) , n ∈ Z (C)  (n π/3) + (-1)n (π/9) , n ∈ Z  (D) (n π/3) + ( -1)n (2 π/9) , n ∈ Z

Tardigrade Admin · Accepted Answer

Given, 4( sin 2 x sin 4 x+ sin 2 x)=3 ⇒ 2( cos 2 x- cos 6 x)+2(1- cos 2 x)=3 ⇒ 2 cos 2 x-2 cos 6 x+2-2 cos 2 x=3 ⇒ -2 cos 6 x+2=3 ⇒ -2 cos 6 x=1 ⇒ cos 6 x=-(1/2) ∴ 6 x=2 n π ± (2 π/3) ⇒ x=(2 n π/6) ± (2/6) × (π/3) =(n π/3) ± (π/9)

Q. If $4 (sin 2 x sin 4 x + sin^{2} x) = 3$ then $x =$

$\frac{2 nπ}{3} \pm \frac{π}{9}, n \in Z$

$\frac{nπ}{3} \pm \frac{π}{9}, n \in Z$

$\frac{nπ}{3} + (- 1)^{n} \frac{π}{9}, n \in Z$

$\frac{nπ}{3} + (- 1)^{n} \frac{2 π}{9}, n \in Z$

Q. If 4(sin2xsin4x+sin2x)=3 then x=

32nπ​±9π​,n∈Z

3nπ​±9π​,n∈Z

3nπ​+(−1)n9π​,n∈Z

3nπ​+(−1)n92π​,n∈Z

Given, 4(sin2xsin4x+sin2x)=3 ⇒2(cos2x−cos6x)+2(1−cos2x)=3 ⇒2cos2x−2cos6x+2−2cos2x=3 ⇒−2cos6x+2=3 ⇒−2cos6x=1 ⇒cos6x=−21​ ∴6x=2nπ±32π​ ⇒x=62nπ​±62​×3π​ =3nπ​±9π​

Q. If $4 (sin 2 x sin 4 x + sin^{2} x) = 3$ then $x =$

$\frac{2 nπ}{3} \pm \frac{π}{9}, n \in Z$

$\frac{nπ}{3} \pm \frac{π}{9}, n \in Z$

$\frac{nπ}{3} + (- 1)^{n} \frac{π}{9}, n \in Z$

$\frac{nπ}{3} + (- 1)^{n} \frac{2 π}{9}, n \in Z$