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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
The general solution of $\sin x-\cos x=\sqrt{2}$, for any integer n is
  • A
    $x= n \pi$
  • ✓
    $x=2 n \pi+\frac{3 \pi}{4}$
  • C
    $x=2 n \pi$
  • D
    $x=(2 n+1) \pi$
✓

Answer

Correct option: B.
$x=2 n \pi+\frac{3 \pi}{4}$
(B) $\sin x-\cos x=\sqrt{2}$
$\Rightarrow \sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}=1$
$\begin{array}{l}\Rightarrow \cos \left(x+\frac{\pi}{4}\right)=-1=\cos \pi \\ \Rightarrow x+\frac{\pi}{4}=2 n \pi \pm \pi \\ \Rightarrow x=2 n \pi+\frac{3 \pi}{4} \text { or } 2 n \pi-\frac{5 \pi}{4}\end{array}$

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