The general solution of sin x - cos x - -2- for any integer n is

Given that, sin x – cos x = √(2) ⇒1/√(2)sin x 1/√(2)cos x = 1 ⇒cos ( x + π/4 ) = 1 ⇒cos ( x + π/4 ) = cos(π) ⇒ x + π/4 = 2nπ±π ⇒ x = 2nπ + 3π/4, 2nπ 5π ...

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Standard XIII

Mathematics

Equations of the Form acosθ+bsinθ = c

The general s...

Question

The general solution of $sin x - cos x = \sqrt{2},$ for any integer n is

n π

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2 n π + \frac{3 π}{4}

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2 n π

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(2 n + 1) π

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Solution

The correct option is B $2 n π + \frac{3 π}{4}$
Given that, $sin x - cos x = \sqrt{2}$
$\Rightarrow \frac{1}{\sqrt{2}} sin x - \frac{1}{\sqrt{2}} cos x = 1$
$\Rightarrow cos (x + \frac{π}{4}) = - 1$
$\Rightarrow cos (x + \frac{π}{4}) = cos (π)$
$\Rightarrow x + \frac{π}{4} = 2 n π \pm π$
$\Rightarrow x = 2 n π + \frac{3 π}{4}, 2 n π - \frac{5 π}{4}$

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The general solution of sinx−cosx=√2, for any integer n is

The correct option is B 2nπ+3π4 Given that, sinx–cosx=√2 ⇒1√2sinx−1√2cosx=1 ​⇒cos(x+π4)=−1 ​⇒cos(x+π4)=cos(π) ​⇒x+π4=2nπ±π ​⇒x=2nπ+3π4,2nπ−5π4

The general solution of $sin x - cos x = \sqrt{2},$ for any integer n is