If cos x = |sin x| then, the general solution is

Question

If cos x = |sin x| then, the general solution is (A) x = n π + (-1)n (π/4) , n ∈ Z  (B) x = n π ± (π/4) , n ∈ Z  (C) x = (2n - 1) π ± (π/4) , n ∈ Z  (D) x = 2n π ± (π/4) , n ∈ Z

Tardigrade Admin · Accepted Answer

cos x=|sin x| ⇒ ± cos x=sin x ⇒ tan x=±1 x=nπ ±(π/4), n∈ Z, but cos x is positive so x =2n π±(π/4), n∈ Z.

Q. If cos x = |sin x| then, the general solution is

$x = nπ + (- 1)^{n} \frac{π}{4}, n \in Z$

$x = nπ \pm \frac{π}{4}, n \in Z$

$x = (2 n - 1) π \pm \frac{π}{4}, n \in Z$

$x = 2 nπ \pm \frac{π}{4}, n \in Z$

$cos x = ∣ s in x ∣$
$\Rightarrow \pm cos x = s in x$
$\Rightarrow t an x = \pm 1$
$x = nπ \pm \frac{π}{4}, n \in Z, b u t cos x i s p os i t i v e so x = 2 nπ \pm \frac{π}{4}, n \in Z .$

Q. If cos x = |sin x| then, the general solution is

x=nπ+(−1)n4π​,n∈Z

x=nπ±4π​,n∈Z

x=(2n−1)π±4π​,n∈Z

x=2nπ±4π​,n∈Z

cosx=∣sinx∣ ⇒±cosx=sinx ⇒tanx=±1 x=nπ±4π​,n∈Z,butcosxispositivesox=2nπ±4π​,n∈Z.

$x = nπ + (- 1)^{n} \frac{π}{4}, n \in Z$

$x = nπ \pm \frac{π}{4}, n \in Z$

$x = (2 n - 1) π \pm \frac{π}{4}, n \in Z$

$x = 2 nπ \pm \frac{π}{4}, n \in Z$

$cos x = ∣ s in x ∣$
$\Rightarrow \pm cos x = s in x$
$\Rightarrow t an x = \pm 1$
$x = nπ \pm \frac{π}{4}, n \in Z, b u t cos x i s p os i t i v e so x = 2 nπ \pm \frac{π}{4}, n \in Z .$