Trigonometric Equations Question 26
Question: If $ 2{{\cos }^{2}}x+3\sin x-3=0,0\le x\le 180^{o} $ , then x =
[MP PET 1986]
Options:
A) $ 30^{o},90^{o},150^{o} $
B) $ 60^{o},120^{o},180^{o} $
C) $ 0^{o},30^{o},150^{o} $
D) $ 45^{o},90^{o},135^{o} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 2-2{{\sin }^{2}}x+3\sin x-3=0 $
$ \Rightarrow $ $ (2\sin x-1),(\sin x-1)=0 $
$ \Rightarrow $ $ \sin x=\frac{1}{2} $ or $ \sin x=1 $
$ \Rightarrow $ $ x=\frac{\pi }{6},,\frac{5\pi }{6} $ , $ \frac{\pi }{2} $ i.e., $ 30{}^\circ ,\text{ 150}{}^\circ \text{, 90}{}^\circ $ .
BETA
