JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Trigonometric Equations

The number of distinct solutions of the equation

\frac{5}{4} cos^{2} 2 x + cos^{4} x + sin^{4} x + cos^{6} x + sin^{6} x = 2

in the interval

[0, 2 π]

Answer: 8

Solution

Simplify the given equation using trigonometric identities. Note that

cos^{4} x + sin^{4} x = 1 - \frac{1}{2} sin^{2} 2 x

and

cos^{6} x + sin^{6} x = 1 - \frac{3}{4} sin^{2} 2 x

. Substituting:

\frac{5}{4} cos^{2} 2 x + 1 - \frac{1}{2} sin^{2} 2 x + 1 - \frac{3}{4} sin^{2} 2 x = 2

\frac{5}{4} cos^{2} 2 x - \frac{5}{4} sin^{2} 2 x = 0

cos^{2} 2 x = sin^{2} 2 x

tan^{2} 2 x = 1

2 x = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{9 π}{4}, \frac{11 π}{4}, \frac{13 π}{4}, \frac{15 π}{4}

Dividing by 2:

x = \frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}, \frac{9 π}{8}, \frac{11 π}{8}, \frac{13 π}{8}, \frac{15 π}{8}

.
All 8 values lie in

[0, 2 π]

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer