cos-2x-1-sin-2x-2-sin-x-cos-x-x-0-2pi-

Question Number 85349 by john santu last updated on 21/Mar/20

\sqrt{\cos 2 x} + \sqrt{1 + \sin 2 x} = 2 \sqrt{\sin x + \cos x}

x \in (0, 2 π)

Commented by john santu last updated on 21/Mar/20

\sqrt{\cos^{2} x - \sin^{2} x} + \sqrt{\sin^{2} x + 2 \sin x \cos x + \cos^{2} x}

= 2 \sqrt{\sin x + \cos x}

\sqrt{(\cos x - \sin x) (\cos x + \sin x)} + \sqrt{{(\sin x + \cos x)}^{2}}

= 2 \sqrt{\sin x + \cos x}

{\begin{cases} (\cos x + \sin x) (\cos x - \sin x) ⩾ 0 \\ \sin x + \cos x ⩾ 0 \end{cases}

(i) \sin x + \cos x = 0 \Rightarrow \tan x = - 1

(i i) \cos x - \sin x = 0 \Rightarrow \tan x = 1

{\frac{3 π}{4} + k π} \cup {- \frac{π}{4} + π n; \frac{π}{4} + π n}

∴ x = - \frac{π}{4} + π n; 2 π k; k, n \in Z