JEE Main 2021 (Online) 25th July Morning ShiftQuestion 1
The sum of all values of $x$ in $[0,2\pi]$, for which $\sin x+\sin2x+\sin3x+\sin4x=0$, is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Pair $\sin x$ with $\sin4x$ and $\sin2x$ with $\sin3x$; a common factor $\sin\frac{5x}2$ appears. Formulas used. (1) $\sin A+\sin B=2\sin\frac{A+B}2\cos\frac{A-B}2$, $\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$Step 1 - Factorise. By (1), $$2\sin\tfrac{5x}2\left(\cos\tfrac{3x}2+\cos\tfrac x2\right)=4\sin\tfrac{5x}{2}\cos x\cos\tfrac x2=0.$$ Step 2 - $\sin\frac{5x}2=0$. $x=0,\frac{2\pi}5,\frac{4\pi}5,\frac{6\pi}5,\frac{8\pi}5,2\pi$; sum $6\pi$. Step 3 - $\cos x=0$. $\frac\pi2,\frac{3\pi}2$; sum $2\pi$. Step 4 - $\cos\frac x2=0$. $x=\pi$. Step 5 - Total. $6\pi+2\pi+\pi=9\pi$. Answer: (B)