The number of solutions of the equation \(4 \sin ^{2} x-4 \cos ^{3} \mathrm{x}+9-4 \cos \mathrm{x}=0 ; \mathrm{x} \in[-2 \pi, 2 \pi]\) is :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Trigonometric Equations

The number of solutions of the equation

4 sin^{2} x - 4 cos^{3} x + 9 - 4 cos x = 0; x \in [- 2 π, 2 π]

is :

Answer: (D)

Using

sin^{2} x = 1 - cos^{2} x

4 (1 - cos^{2} x) - 4 cos^{3} x + 9 - 4 cos x = 0

- 4 cos^{3} x - 4 cos^{2} x - 4 cos x + 13 = 0

4 cos^{3} x + 4 cos^{2} x + 4 cos x = 13.

Let

c = cos x \in [- 1, 1]

. The left side,

4 c^{3} + 4 c^{2} + 4 c

, attains its largest value on

[- 1, 1]

c = 1

, where it equals

4 + 4 + 4 = 12

, which is still less than

13

.

Since the left side can never reach

13

for any real

x

, the equation has no solution in

[- 2 π, 2 π]

— or anywhere. The number of solutions is

0

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Single correct option