Question

If $(\sin x+\sin 2x+\sin 3x{)}^2+(\cos x+\cos 2x+\cos 3x{)}^2=1$ then $\cos x$ can have the value to

• A. $1$
• B. $0$
• C. $\frac{-1}{2}$
• D. $-1$

Answer 1

Answer: (B), (D)

The correct options are
B $0$
D $-1$

$\sin x+\sin 2x+\sin 3x$

$=\sin (2x-x)+\sin 2x+\sin (2x+x)$

$=2\sin 2x\cos x+\sin 2x$ [ by using $\sin (A+B)=\sin A\cos B+\sin B\cos A$ and $\sin (A-B)=\sin A\cos B-\sin B\cos A$ ]

$=\sin 2x(2\cos x+1)........\left(i\right)$

$\cos x+\cos 2x+\cos 3x$

$=\cos (2x-x)+\cos 2x+\cos (2x+x)$

$=2\cos 2x\cos x+\cos 2x$ [By using $\cos (a-b)=\cos a\cdot \cos b+\sin a\cdot sinb$ and $\cos (a+b)=\cos a\cdot \cos b-\sin a\cdot sinb$]

$=\cos 2x(2\cos x+1).....\left(ii\right)$

$\therefore (\sin x+\sin 2x+\sin 3x{)}^2+(\cos x+\cos 2x+\cos 3x{)}^2=1$

${\sin }^{2}2x(2\cos x+1{)}^2+{\cos }^{2}2x(2\cos x+1{)}^2=\mathrm{1.......}\left[From\right(i\left)\right(ii\left)\right]$

$\Rightarrow (2\cos x+1{)}^2=1$

$\Rightarrow 2\cos x+1=\pm 1$

$\therefore \cos x=0or-1$