Question

The equation $2\sin \frac{x}{2}{\cos }^{2}x-2\sin \frac{x}{2}{\sin }^{2}x={\cos }^{2}x-{\sin }^{2}x$ has a root for which

• A. $\sin 2x=1$
• B. $\sin 2x=-1$
• C. $\cos x=\frac{1}{2}$
• D. $\cos 2x=-\frac{1}{2}$

Answer 1

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

The correct options are
A $\cos 2x=-\frac{1}{2}$
B $\sin 2x=-1$
C $\cos x=\frac{1}{2}$
D $\sin 2x=1$

The given equation can be written as
$2\sin \left(x/2\right)({\cos }^{2}x-{\sin }^{2}x)={\cos }^{2}x-{\sin }^{2}x$

$\Rightarrow \left(2\sin \right(x/2)-1)\cos 2x=0$

Hence $\cos 2x=0$ or $\sin \left(x/2\right)=1/2$

i.e. $2x=n\pi +\pi /2$ or
$x/2=k\pi +(-1{)}^k(\pi /6\left)\right(n,k\epsilon I)$.

In other words,
$x=n\pi /2+\pi /4$ or $x=2k\pi +(-1{)}^k\pi /3$

If $x=n\pi /2+\pi /4,$ then $\sin 2x=\pm 1$,

and if $x=2k\pi +(-1{)}^k\pi /3,\cos x$

$=\cos \left(\pi /3\right)=1/2$

and $\cos 2x=\cos \left(2\pi /3\right)=-1/2$.