Question

$\sqrt{cos2x}+\sqrt{(1+sin2x)}=2\sqrt{(sinx+cosx)}$, if-

• A. $sinx+cosx=0$
• B. $x=2n\pi$
• C. $x=n\pi -\pi /4$
• D. $sinx-cosx=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $sinx+cosx=0$
B $x=2n\pi$
C $x=n\pi -\pi /4$

$\sqrt{cos2x}+\sqrt{(1+sin2x)}=2\sqrt{(sinx+cosx)}$

$\Rightarrow \sqrt{{\cos }^{2}x-{\sin }^{2}x}+\sqrt{(\cos x+\sin x{)}^2}=2\sqrt{(sinx+cosx)}$

$\Rightarrow \sqrt{(\cos x+\sin x)}(\sqrt{(\cos x-\sin x)}+\sqrt{(\cos x+\sin x)})=2\sqrt{(sinx+cosx)}$

Either $\sqrt{(\cos x+\sin x)}=0or\sqrt{(\cos x-\sin x)}-\sqrt{(\cos x+\sin x)}=2$

$\Rightarrow (\cos x+\sin x)=0$

$\Rightarrow \tan x=-1$

$\Rightarrow x=n\pi -\frac{\pi }{4},n\in I$
or $\sqrt{(\cos x-\sin x)}+\sqrt{(\cos x+\sin x)}=2$

$2\cos x+2\sqrt{{\cos }^{2}x-{\sin }^{2}x}=4$

$\Rightarrow (2\cos x-4{)}^2=4({\cos }^{2}x-{\sin }^{2}x)$

$\Rightarrow {\cos }^{2}x+4\cos x-5=0$

$\Rightarrow (\cos x+5)(\cos x-1)=0$

$\Rightarrow \cos x=-5$(not possible )or $1$

$\Rightarrow \cos x=1$

$\Rightarrow x=2n\pi$