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

Solve the following for x:
sin1(1x)2sin1x=π2

Open in App
Solution

sin1(1x)2sin1x=π2 ....(1)
sin1(1x)=π2+2sin1x
1x=sin[π2+2sin1x]
1x=cos[2sin1x]
1x=12[sin(sin1x)]2[cos(2x)=12sin2x]
1x=12x2
2x2x=0
x(2x1)=0
x=0,x=12
Put x=0 in equation (1)
LHS=sin112sin10=π2=RHS
Put x=12 in equation (1)
LHS=sin1(12)2sin1(12)=π62π6=π6
LHSπ2RHS
So, x=12 is not a solution.
x=0 is a solution to given equation.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 2
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon