JEE Main 2016 (Offline)Question 1
If $0\le x<2\pi$, then the number of real values of $x$, which satisfy the equation $\cos x+\cos2x+\cos3x+\cos4x=0$, is
Trigonometric Ratio & Identities
Solution
Concept. Pair $\cos x$ with $\cos4x$ and $\cos2x$ with $\cos3x$; the sum becomes a product of three cosines. Formulas used. (1) $\cos P+\cos Q=2\cos\dfrac{P+Q}{2}\cos\dfrac{P-Q}{2}$Step 1 - Factorise. By (1), $$\cos x+\cos4x=2\cos\tfrac{5x}{2}\cos\tfrac{3x}{2},\qquad\cos2x+\cos3x=2\cos\tfrac{5x}{2}\cos\tfrac{x}{2}.$$ Adding, and using (1) again on $\cos\frac{3x}2+\cos\frac x2$: $$4\cos\tfrac{5x}{2}\,\cos x\,\cos\tfrac x2=0.$$ Step 2 - Zeros in $[0,2\pi)$. $\cos\frac{5x}2=0$: $x=\frac\pi5,\frac{3\pi}5,\pi,\frac{7\pi}5,\frac{9\pi}5$. $\cos x=0$: $x=\frac\pi2,\frac{3\pi}2$. $\cos\frac x2=0$: $x=\pi$ (already counted). Step 3 - Count. $5+2=7$ distinct values. Answer: (A)