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
71 views
in Mathematics by (92.4k points)
closed by
The number of values of x in the interval `[0,5pi]` satisfying the equation `3sin^2x-7sinx+2=0` is

1 Answer

0 votes
by (90.5k points)
selected by
 
Best answer
Correct Answer - C
Given,`3sin^2x-7sinx+2=0`
`(3sinx-1)(sinx-2)=0`
`rArr sinx=(1)/(3) or 2`
`rArr sinx=(1)/(3) [because sinx=2]`
Let `sin^(-1)(1)/(3)=alpha,0lt alpha lt (pi)/(2)`, then `alpha,pi-alpha,+alpha,3pi-alpha,4pi+alpha,5pi-alpha` are the solution in `[0,5pi]`
Hence,required number of solutions are 6.

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

...