If x1 and x2 are the two solutions of the equation 2 cos x sin 3 x= sin 4 x+1 lying in the interval [0,2 π], then the value of |x1-x2| is

Question

If x1 and x2 are the two solutions of the equation 2 cos x sin 3 x= sin 4 x+1 lying in the interval [0,2 π], then the value of |x1-x2| is (A) (π/4) (B) (π/2) (C) π (D) 2 π

Tardigrade Admin · Accepted Answer

sin 4 x+ sin 2 x= sin 4 x+1 ∴ sin 2 x=1 ⇒ 2 x=(π/2) text or (5 π/2) ∴ x=(π/4), (5 π/4) ∴|x1-x2|=π

Q. If $x_{1}$ and $x_{2}$ are the two solutions of the equation $2 cos x sin 3 x = sin 4 x + 1$ lying in the interval $[0, 2 π]$ , then the value of $∣ x_{1} - x_{2} ∣$ is

$\frac{π}{4}$

$\frac{π}{2}$

$π$

$2 π$

$sin 4 x + sin 2 x = sin 4 x + 1$
$∴ sin 2 x = 1$
$\Rightarrow 2 x = \frac{π}{2} or \frac{5 π}{2}$
$∴ x = \frac{π}{4}, \frac{5 π}{4}$
$∴ ∣ x_{1} - x_{2} ∣ = π$

Q. If x1​ and x2​ are the two solutions of the equation 2cosxsin3x=sin4x+1 lying in the interval [0,2π], then the value of ∣x1​−x2​∣ is

4π​

2π​

π

2π

sin4x+sin2x=sin4x+1 ∴sin2x=1 ⇒2x=2π​ or 25π​ ∴x=4π​,45π​ ∴∣x1​−x2​∣=π

Q. If $x_{1}$ and $x_{2}$ are the two solutions of the equation $2 cos x sin 3 x = sin 4 x + 1$ lying in the interval $[0, 2 π]$ , then the value of $∣ x_{1} - x_{2} ∣$ is

$\frac{π}{4}$

$\frac{π}{2}$

$π$

$2 π$

$sin 4 x + sin 2 x = sin 4 x + 1$
$∴ sin 2 x = 1$
$\Rightarrow 2 x = \frac{π}{2} or \frac{5 π}{2}$
$∴ x = \frac{π}{4}, \frac{5 π}{4}$
$∴ ∣ x_{1} - x_{2} ∣ = π$