If cos x -sin x - then the general solution is A- x-n -1n-4- n - ZB- x-n -4- n - Z- x-2 n-1 -4- n - ZD- x-2 n -4- n - Z

The correct option is B x= n π±π/4, n∈ Z nbsp; nbsp;cos x=|sin x| nbsp; ⇒cos^2 x= sin ^2 x ⇒ 1 sin^2 x= sin^2 x ⇒ 2sin^2 x=1 ⇒sin x = ±1/√(2) ⇒ x= n π±π ...

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Standard XII

Physics

Antiderivative

If cos x =|si...

Question

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

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

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x = n π \pm \frac{π}{4}, n \in Z

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

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x = 2 n π \pm \frac{π}{4}, n \in Z

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Solution

The correct option is B $x = n π \pm \frac{π}{4}, n \in Z$
$cos x = | sin x |$
$\Rightarrow {cos}^{2} x = {sin}^{2} x$
$\Rightarrow 1 - {sin}^{2} x = {sin}^{2} x$
$\Rightarrow 2 {sin}^{2} x = 1 \Rightarrow sin x = \pm \frac{1}{\sqrt{2}}$
$\Rightarrow x = n π \pm \frac{π}{4}, n \in Z$

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If cosx=|sinx|, then the general solution is

The correct option is B x=nπ±π4, n∈Z cosx=|sinx| ⇒cos2x=sin2x ⇒1−sin2x=sin2x ⇒2sin2x=1⇒sinx=±1√2 ⇒x=nπ±π4, n∈Z

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

The correct option is B $x = n π \pm \frac{π}{4}, n \in Z$
$cos x = | sin x |$
$\Rightarrow {cos}^{2} x = {sin}^{2} x$
$\Rightarrow 1 - {sin}^{2} x = {sin}^{2} x$
$\Rightarrow 2 {sin}^{2} x = 1 \Rightarrow sin x = \pm \frac{1}{\sqrt{2}}$
$\Rightarrow x = n π \pm \frac{π}{4}, n \in Z$