Question

The solution of the equation $1+(\sin x-\cos x)\sin \frac{\pi }{4}=2{\cos }^2\frac{5x}{2}$ is / are

• A. $x=\frac{n\pi }{3}+\frac{\pi }{48},nϵZ$
• B. $x=\frac{n\pi }{2}+\frac{5\pi }{16},nϵZ$
• C. $x=\frac{n\pi }{3}+\frac{\pi }{4},nϵZ$
• D. $x=\frac{n\pi }{2}+\frac{7\pi }{8},nϵZ$

Answer 1

Answer: (A), (B)

$x=\frac{n\pi }{2}+\frac{5\pi }{16},nϵZ$
$x=\frac{n\pi }{3}+\frac{\pi }{48},nϵZ$

$1+\frac{1}{\sqrt{2}}\left(\sin x-\cos x\right)=2{\cos }^2\frac{5x}{2}$
$\Rightarrow -\cos (x+\frac{\pi }{4})=\cos 5x$
$\Rightarrow -x-\frac{\pi }{4}=2n\pi \pm 5x$
$\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{16},\frac{n\pi }{3}+\frac{\pi }{48}$