Trigonometric Equations Question 40
Question: The number of values of x in the interval
[0, 5 $ \pi $ ] satisfying the equation $ 3{{\sin }^{2}}x-7\sin x+2=0 $ is [IIT 1998; MP PET 2000; Pb. CET 2003]
Options:
A) 0
B) 5
C) 6
D) 10
Show Answer
Answer:
Correct Answer: C
Solution:
- $ 3{{\sin }^{2}}x-7\sin x+2=0 $
$ \Rightarrow $ $ 3{{\sin }^{2}}x-6\sin x-\sin x+2=0 $
$ \Rightarrow $ $ 3\sin (\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} $ , ( $ \because \sin x\ne 2 $ ) Let $ {{\sin }^{-1}}\frac{1}{3}=\alpha $ , $ 0<\alpha <\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 $ Required number of solutions = 6.
BETA
