Question

solve the trigonomic equation where 0 is less than or equal to x which is less than 2pi : sin^2 2x+3sin2x+2=0

Answer 1

let u=sin(2x)

Answer 2

u^2+3u+2=0

Answer 3

(sin2x+1)(sin2x+2)=0
sin2x=-1 only possible
s0 x=--pi

Answer 4

(u+1)(u+2)=0
u=-1
u=-2

Answer 5

but u=sin(2x)
so sin(2x)=-1
and
sin(2x)=-2
and sin stays between -1 and 1
so yes the second one isn't possible good job dipan

Answer 6

thanks

Answer 7

remember sin(3pi/2)=-1
so we need 2x=3pi/2
just solve for x

Answer 8

x=3pi/4