Question

Solve $\sin x+\sin \left(\frac{\pi }{2}\sqrt{(1-\cos 2x{)}^2+{\sin }^{2}2x}\right)=0,x\in [\frac{5\pi }{2},\frac{7\pi }{2}]$

• A. $x=8\pi /3$
• B. $x=10\pi /3$
• C. $x=13\pi /4$
• D. $x=5\pi /2$

Answer 1

Answer: (C)

$x=13\pi /4$

$\sin x+\sin \frac{\pi }{8}\sqrt{{(1-\cos x)}^2+{\sin }^{2}x}$

$\Rightarrow {\sin }^{2}x={\sin }^2\left(\frac{\pi }{8}\right){(1-\cos x)}^2+{\sin }^{2}x={\sin }^2\left(\frac{\pi }{8}\right)(2-2\cos x)$

$\Rightarrow 1-{\cos }^{2}x=(1-\cos \frac{\pi }{8})(1-\cos x)$

$\Rightarrow (1-\cos x)[1+\cos x-(1-\cos \frac{\pi }{4})]=0$

$\Rightarrow (1-\cos x)\left(\cos x+\cos \frac{\pi }{8}\right)=0$

$\Rightarrow \cos x=1$ or $\cos x=-\frac{1}{2}\Rightarrow x=2n\pi ,x=2m\pi \pm \frac{3\pi }{4}$

$x=2n\pi$ givens no solution between $\frac{5\pi }{2}$ and $\frac{7\pi }{2}$
and $x=2m\pi \pm \frac{3\pi }{4}=(8m\pm 3)\frac{\pi }{4}$

Gives solution $2\pi +\frac{3\pi }{4}$ and

$4\pi -\frac{3\pi }{4}$

i.e $\frac{11\pi }{4}$ and $\frac{13\pi }{2}$

in the interval $\frac{5\pi }{2}$ and $\frac{7\pi }{2}$

But $\frac{11\pi }{4}$ does not satisfies the given equation

Hence the solution in the given interval is $x=\frac{13\pi }{4}$