Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
561 views
in Trigonometry by (22.8k points)
closed by

The general solution of sin 3x + sin x – 3 sin 2x = cos 3x + cos x – 3 cos 2x is

(a)  \(\frac{nπ}{2}+ \frac{π}{8}\) ,  n ∈ I

(b)  \(\frac{nπ}{2}- \frac{π}{8}\) ,  n ∈ I 

(c)  \(nπ+ \frac{π}{8}\) ,  n ∈ I

(d)  \(nπ- \frac{π}{8}\) ,  n ∈ I

1 Answer

+1 vote
by (23.7k points)
selected by
 
Best answer

Answer : (a) \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

(sin 3x + sin x) – 3 sin 2x = (cos 3x + cos x) – 3 cos 2x 

⇒ 2 sin 2x cos x – 3 sin 2x = 2 cos 2x . cos x – 3 cos 2x 

⇒ sin 2x (2 cos x – 3) = cos 2x (2 cos x – 3) 

⇒ (2 cos x – 3) (sin 2x – cos 2x) = 0 

⇒ (2 cos x – 3) = 0 or (sin 2x – cos 2x) = 0 

⇒ cos x = \(\frac{3}{2}\) or sin 2x = cos 2x 

∵  – 1 ≤ cos x ≤ 1, cos x ≠  \(\frac{3}{2}\)

∴ sin 2x = cos 2x ⇒ tan 2x = 1 

⇒ tan 2x = tan (π/4) 

⇒ 2x = nπ + π/4

x = \(\frac{nπ}{2}+\frac{π}{8}\), n ∈ I

No related questions found

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

...