sin2x+cosx=0 ........ find general solution.
Answers
Answered by
50
sin(2x) - cosx = 0
2sinxcosx - cosx = 0
cosx(2sinx - 1) = 0
cosx = 0 or sinx = 1/2
x = pi/2, 3pi/2, pi/2+2kpi, 3pi/2+2kpi
x= pi/6, 5pi/6, pi/6+2kpi, 5pi/6+2kpi where k is any integer
HOPE IT HELPS U....!!❤❤❤❤❤❤❤❤❤
2sinxcosx - cosx = 0
cosx(2sinx - 1) = 0
cosx = 0 or sinx = 1/2
x = pi/2, 3pi/2, pi/2+2kpi, 3pi/2+2kpi
x= pi/6, 5pi/6, pi/6+2kpi, 5pi/6+2kpi where k is any integer
HOPE IT HELPS U....!!❤❤❤❤❤❤❤❤❤
ira96:
ya
hii all of u
guys in this sum the value of sinx=-1/2
how come it is correct
if 2sinx+1=0
then it is -1/2
Answered by
28
sin2x+cosx=0
sin2x=2sinx×cosx
2sinx×cosx+cosx=0
cosx(2sinx +1)=0
cosx=0 2sinx+1=0
sinx=-1/2
let sinx=siny..........1
sinx=-1/2........2
as we know sin 30=1/2
siny=7pi/6
hence x=npi+(-1)^n×7pi/6where n belongs to z
hence for cosx=0
x=(2n+1) pi/2 same n belongs to z
or
sinx=-1/2
npi+(-1)^n×7pi/6
sin2x=2sinx×cosx
2sinx×cosx+cosx=0
cosx(2sinx +1)=0
cosx=0 2sinx+1=0
sinx=-1/2
let sinx=siny..........1
sinx=-1/2........2
as we know sin 30=1/2
siny=7pi/6
hence x=npi+(-1)^n×7pi/6where n belongs to z
hence for cosx=0
x=(2n+1) pi/2 same n belongs to z
or
sinx=-1/2
npi+(-1)^n×7pi/6
hope it will help u out
Similar questions