MHT CET · Maths · Trigonometric Equations
The number of values of \(x\) in the interval \((0,5 \pi)\) satisfying the equation \(3 \sin ^2 x-7 \sin x+2=0\)
- A 0
- B 5
- C 6
- D 10
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& 3 \sin ^2 x-7 \sin x+2=0 \\
& \Rightarrow 3 \sin ^2 x-6 \sin x-\sin x+2=0 \\
& \Rightarrow 3 \sin x(\sin x-2)-(\sin x-2)=0 \\
& \Rightarrow(3 \sin x-1)(\sin x-2)=0 \\
& \Rightarrow \sin x=\frac{1}{3} \text { or } 2 \\
& \Rightarrow \sin x=\frac{1}{3}
\end{aligned} \ldots .[\because \sin x \neq 2]\)
Let \(\sin ^{-1} \frac{1}{3}=\alpha, 0 \lt \alpha \lt \frac{\pi}{2}\) are the solutions in \([0,5 \pi]\). Then, \(\alpha, \pi-\alpha, 2 \pi+\alpha, 3 \pi-\alpha\), \(4 \pi+\alpha, 5 \pi-\alpha\) are the solutions in \([0,5 \pi]\).
\(\therefore \quad\) number of solutions \(=6\)
& 3 \sin ^2 x-7 \sin x+2=0 \\
& \Rightarrow 3 \sin ^2 x-6 \sin x-\sin x+2=0 \\
& \Rightarrow 3 \sin x(\sin x-2)-(\sin x-2)=0 \\
& \Rightarrow(3 \sin x-1)(\sin x-2)=0 \\
& \Rightarrow \sin x=\frac{1}{3} \text { or } 2 \\
& \Rightarrow \sin x=\frac{1}{3}
\end{aligned} \ldots .[\because \sin x \neq 2]\)
Let \(\sin ^{-1} \frac{1}{3}=\alpha, 0 \lt \alpha \lt \frac{\pi}{2}\) are the solutions in \([0,5 \pi]\). Then, \(\alpha, \pi-\alpha, 2 \pi+\alpha, 3 \pi-\alpha\), \(4 \pi+\alpha, 5 \pi-\alpha\) are the solutions in \([0,5 \pi]\).
\(\therefore \quad\) number of solutions \(=6\)
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