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Trigonometrical Equations
hard
समीकरण $4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0$; $x \in[-2 \pi, 2 \pi]$ के हलों की संख्या है :
A
$1$
B
$3$
C
$2$
D
$0$
(JEE MAIN-2024)
Solution
$ 4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi] $
$ 4-4 \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 $
$ \text { L.H.S. } \leq 12 \text { can't be equal to } 13 .$
Standard 11
Mathematics
