Q89 / 180 · COMEDK 2024 (Memory Based) · Mathematics
The number of solution of $\cos ^2 x=3-3 \sin x$ $\forall x \in[0,2 \pi]$ is
- A0
- Correct answerB1
- C2
- D3
Solution
We have, $\cos ^2 x=3-3 \sin x$
$\begin{aligned}<br/>& 1-\sin ^2 x=3-3 \sin x \\<br/>& \sin ^2 x-3 \sin x+2=0 \\<br/>& (\sin x-1)(\sin x-2)=0 \\<br/>& \therefore \quad \sin x-2=0 \\<br/>& \sin x=2 \text { has no solution }<br/>\end{aligned}$
$\therefore \sin x=1$ has one solution which is $x=\frac{\pi}{2}$
$\begin{aligned}<br/>& 1-\sin ^2 x=3-3 \sin x \\<br/>& \sin ^2 x-3 \sin x+2=0 \\<br/>& (\sin x-1)(\sin x-2)=0 \\<br/>& \therefore \quad \sin x-2=0 \\<br/>& \sin x=2 \text { has no solution }<br/>\end{aligned}$
$\therefore \sin x=1$ has one solution which is $x=\frac{\pi}{2}$
Hence, the correct answer is option: 'B'.