1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If sin−1(1−x)−2sin−1x=π2, then x is equal to

A
0,12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1,12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 0sin−1(1−x)−2sin−1x=π2−2sin−1x=π2−sin−1(1−x)We know that sin−1x+cos−1x=π2Replace x by 1−xsin−1(1−x)+cos−1(1−x)=π2−2sin−1x=cos−1(1−x) .....(1)Let sin−1x=a,Hence our equation becomes−2a=cos−1(1−x)⇒cos2a=1−x⇒1−2sin2a=1−x⇒1−x=1−2sin2a⇒1−x=1−2x2⇒−2x2+x=0⇒−x(2x−1)=0⇒x=0,x=12But x=12 does not satisfy the equationTaking equation sin−1(1−x)−2sin−1x=π2Put x=12 in L.H.Ssin−1(1−12)−2sin−1(12)=sin−1(12)−2sin−1(12)=−sin−1(12)=−π6≠π2Hence x=12 not possible.∴x=0 is the only solution.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Property 4
MATHEMATICS
Watch in App
Join BYJU'S Learning Program