JEE Main 2024 (Online) 1st February Evening ShiftQuestion 1
The number of solutions of the equation $4\sin^{2}x-4\cos^{3}x+9-4\cos x=0$; $x\in[-2\pi,2\pi]$ is :
Trigonometric Ratio & Identities
Solution
Concept. In $c=\cos x$ the equation is a cubic that is increasing and still negative at $c=1$. Formulas used. (1) $\sin^2x=1-\cos^2x$Step 1 - Cubic. By (1), $4(1-c^2)-4c^3+9-4c=0$, i.e. $$f(c)=4c^3+4c^2+4c-13=0.$$ Step 2 - Increasing. $f'(c)=12c^2+8c+4>0$ (its discriminant $64-192<0$). Step 3 - No root in $[-1,1]$. $f(1)=4+4+4-13=-1<0$, so $f(c)<0$ for every $c\le1$. Step 4 - Count. There are $0$ solutions. Answer: (A)