find the general solution of 2cos^2x+3sinx=0
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2 (1-sin^2x) + 3sinx=0
2-2sin^2x + 3sinx=0
2sin^2x -3sinx -2=0
let y = sind
then. 2y^2 -3y -2=0
2y^2 -4y+y-2=0
2y(y-2) +(y-2)=0
(2y +1) (y-2)=0
y= -1/2 and y= 2
therefore
sinx = -1/2.
Sin x is negative in IIIrd and IVth quadrant. Since sin π6=12, sin x would be −12 for x= π+π6and 2π−π6.
The Principal solution is this x= 7π6and11π6
sinx =2
x= sin ^(-1) 2= 1.5708 ± 1.31696 i
hope it's correct!!
if so or not reply me!
2-2sin^2x + 3sinx=0
2sin^2x -3sinx -2=0
let y = sind
then. 2y^2 -3y -2=0
2y^2 -4y+y-2=0
2y(y-2) +(y-2)=0
(2y +1) (y-2)=0
y= -1/2 and y= 2
therefore
sinx = -1/2.
Sin x is negative in IIIrd and IVth quadrant. Since sin π6=12, sin x would be −12 for x= π+π6and 2π−π6.
The Principal solution is this x= 7π6and11π6
sinx =2
x= sin ^(-1) 2= 1.5708 ± 1.31696 i
hope it's correct!!
if so or not reply me!
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