SATHEE: Trigonometrical Equations 2 Question 2

Trigonometrical Equations 2 Question 2

2. The set of values of $θ$ satisfying the inequation $2 \sin^{2} θ - 5 \sin θ + 2 > 0$ , where $0 < θ < 2 π$ , is

(2006, 3M)

(a) $0, \frac{π}{6} \cup \frac{5 π}{6}, 2 π$

(b) $0, \frac{π}{6} \cup \frac{5 π}{6}, 2 π$

(d) None of the above

Show Answer

Answer:

Correct Answer: 2. (a)

Solution:

Since, $2 \sin^{2} θ - 5 \sin θ + 2 > 0$

$\Rightarrow (2 \sin θ - 1) (\sin θ - 2) > 0$

[where, $(\sin θ - 2) < 0, \forall θ \in R$ ]

$∴ (2 \sin θ - 1) < 0$

$\Rightarrow \sin θ < \frac{1}{2}$

$∴$ From the graph, $θ \in 0, \frac{π}{6} \cup \frac{5 π}{6}, 2 π$

पिछला

अगला

गोपनीयता नीति

द्वारा संचालित Prutor@IITK

sathee@iitk.ac.in

SATHEE का मीडिया कवरेज

खेल स्टोर

ऐप स्टोर

Ask SATHEE

Welcome to SATHEE !
Select from 'Menu' to explore our services, or ask SATHEE to get started. Let's embark on this journey of growth together! 🌐📚🚀🎓

I'm relatively new and can sometimes make mistakes.
If you notice any error, such as an incorrect solution, please use the thumbs down icon to aid my learning.
To begin your journey now, click on "I understand".

Trigonometrical Equations 2 Question 2

2. The set of values of θ satisfying the inequation 2sin2⁡θ−5sin⁡θ+2>0, where 0<θ<2π, is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Menu

2. The set of values of $θ$ satisfying the inequation $2 \sin^{2} θ - 5 \sin θ + 2 > 0$ , where $0 < θ < 2 π$ , is