Use app×
Ask a Question
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE

If sin^2 θ – 2sinθ – 1 = 0 is to be satisfied for exactly 4 distinct values of θ ∈ [0, nπ], ∀ n∈N ; then the least value of n is

← Prev Question Next Question →
0 votes
179 views
in Trigonometry by (53.5k points)
closed by

If sin2θ – 2sinθ – 1 = 0 is to be satisfied for exactly 4 distinct values of θ ∈ [0, nπ], ∀ n∈N ; then the least value of n is

(a) 2

(b) 6

(c) 4

(d) 8

Challenge Your Friends with Exciting Quiz Games – Click to Play Now!

1 Answer

+1 vote
by (53.7k points)
selected by
 
Best answer

Correct option is (c) 4

clearly four solutions lie in [0,4π] 

\(\therefore\) least value of n is 4

Also 5th solution lies in [0,5π] 

\(\therefore\) greatest value can be 5.

← Prev Question Next Question →

Find MCQs & Mock Test

  1. JEE Main 2026 Test Series
  2. NEET Test Series
  3. Class 12 Chapterwise MCQ Test
  4. Class 11 Chapterwise Practice Test
  5. Class 10 Chapterwise MCQ Test
  6. Class 9 Chapterwise MCQ Test
  7. Class 8 Chapterwise MCQ Test
  8. Class 7 Chapterwise MCQ Test

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

User Contribution to Sarthaks eConnect
Managed by Sarthaks Digitizing India
...