Question

Solve the following equations:
(a) $\sin x+\sqrt{2}=\cos x$
(b) $\csc \theta =1+\cot \theta$

• A. (a) $x=2n\pi -\frac{\pi }{4},n\in I$
  (b) $m\pi +\frac{\pi }{2}.m\in I$
• B. (a) $x=n\pi -\frac{\pi }{2},n\in I$
  (b) $m\pi +\frac{\pi }{2}.m\in I$
• C. (a) $x=2n\pi -\frac{\pi }{4},n\in I$
  (b) $2m\pi +\frac{\pi }{2}.m\in I$
• D. (a) $x=n\pi -\frac{\pi }{6},n\in I$
  (b) $2m\pi +\frac{\pi }{4}.m\in I$

Answer 1

The correct option is C (a) $x=2n\pi -\frac{\pi }{4},n\in I$

(b) $2m\pi +\frac{\pi }{2}.m\in I$
a) $\sin x+\sqrt{2}=\cos x\Rightarrow$$\sin x-\cos x=-\sqrt{2}$

$\Rightarrow \frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=-1$

$\Rightarrow \sin \frac{\pi }{4}\sin x-\cos \frac{\pi }{4}\cos x=-1$

$\Rightarrow \cos \frac{\pi }{4}\cos x-\sin \frac{\pi }{4}\sin x=1$

$\Rightarrow \cos (x+\frac{\pi }{4})=1\Rightarrow x+\frac{\pi }{4}=2n\pi$

$\Rightarrow x=2n\pi -\frac{\pi }{4}$
b) $\csc \theta =1+\cot \theta \Rightarrow \frac{1}{\sin \theta }=1+\frac{\cos \theta }{\sin \theta }$

$\Rightarrow \sin \theta +\cos \theta =1$

$\Rightarrow \frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{1}{\sqrt{2}}=\cos (\theta -\frac{\pi }{4})$

$\Rightarrow \theta -\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$

$\Rightarrow \theta -\frac{\pi }{4}=2n\pi -\frac{\pi }{4}$ and $\Rightarrow \theta -\frac{\pi }{4}=2n\pi +\frac{\pi }{4}$

$\because \sin \theta \ne 0\Rightarrow \theta \ne n\pi \therefore \theta =2m\pi +\frac{\pi }{2}$