The number of solution of the equation operatornamesgn( sin x)- operatornamesgn( sin 2 x)= sin 2 x+2 sin x in [(-5 π/2), (7 π/2)] is [Note: operatornamesgn( k ) denotes signum function of k.]

Question

The number of solution of the equation operatornamesgn( sin x)- operatornamesgn( sin 2 x)= sin 2 x+2 sin x in [(-5 π/2), (7 π/2)] is [Note: operatornamesgn( k ) denotes signum function of k.] (A) 10 (B) 6 (C) 13 (D) 9

Tardigrade Admin · Accepted Answer

text Case-I: sin x=0 ⇒ x=-2 π,-π, 0, π, 2 π, 3 π operatornamesgn( sin x)- operatornamesgn( sin 2 x)= sin 2 x+2 sin x text has 6 text roots text Case-II: sin x ≠ 0 operatornamesgn( sin x)-1= sin 2 x+2 sin x operatornamesgn( sin x)=(1+ sin x)2 text no. real roots.

Q. The number of solution of the equation $sgn (sin x) - sgn (sin^{2} x) = sin^{2} x + 2 sin x$ in $[\frac{- 5 π}{2}, \frac{7 π}{2}]$ is [Note: $sgn (k)$ denotes signum function of $k$ .]

10

6

13

9

$Case-I : sin x = 0 \Rightarrow x = - 2 π, - π, 0, π, 2 π, 3 π$
$sgn (sin x) - sgn (sin^{2} x) = sin^{2} x + 2 sin x$
$has 6 roots$
$Case-II : sin x \neq = 0$
$sgn (sin x) - 1 = sin^{2} x + 2 sin x$
$sgn (sin x) = (1 + sin x)^{2}$
$no. real roots.$

Q. The number of solution of the equation sgn(sinx)−sgn(sin2x)=sin2x+2sinx in [2−5π​,27π​] is [Note: sgn(k) denotes signum function of k.]

10

6

13

9

Case-I : sinx=0⇒x=−2π,−π,0,π,2π,3π sgn(sinx)−sgn(sin2x)=sin2x+2sinx has 6 roots Case-II : sinx=0 sgn(sinx)−1=sin2x+2sinx sgn(sinx)=(1+sinx)2 no. real roots.

Q. The number of solution of the equation $sgn (sin x) - sgn (sin^{2} x) = sin^{2} x + 2 sin x$ in $[\frac{- 5 π}{2}, \frac{7 π}{2}]$ is [Note: $sgn (k)$ denotes signum function of $k$ .]

$Case-I : sin x = 0 \Rightarrow x = - 2 π, - π, 0, π, 2 π, 3 π$
$sgn (sin x) - sgn (sin^{2} x) = sin^{2} x + 2 sin x$
$has 6 roots$
$Case-II : sin x \neq = 0$
$sgn (sin x) - 1 = sin^{2} x + 2 sin x$
$sgn (sin x) = (1 + sin x)^{2}$
$no. real roots.$