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Trigonometrical Equations
easy
કોઇ $n$ પૂર્ણાક માટે $\sin x - \cos x = \sqrt 2 $ નો વ્યાપક ઉકેલ મેળવો.
A
$n\pi $
B
$2n\pi + \frac{{3\pi }}{4}$
C
$2n\pi $
D
$(2n + 1)\,\pi $
Solution
(b) $\sin x\frac{1}{{\sqrt 2 }} – \cos x\frac{1}{{\sqrt 2 }} = 1$
$\Rightarrow \cos \left( {x + \frac{\pi }{4}} \right) = – 1$
==> $x + \frac{\pi }{4} = 2n\pi \pm \pi $
$\Rightarrow 2n\pi + \frac{{3\pi }}{4}{\rm{\,\,or\,\,}}\,\,2n\pi – \frac{{5\pi }}{4}$ .
Standard 11
Mathematics
