how do you solve 2sin^2(3x)-5sin3x-3=0 for [0,2pi]?

Question

thanks!

Answer 1

2sin^2(3x)-5sin3x-3=0

(2sin3x + 1) (sin3x - 3) = 0

2sin3x = 0
sin3x = 0
x = 0, pi, pi/3, 2pi/3 2pi

sin3x -3 = 0
sin3x = 3


the possible answers are 0, pi, pi/3, 2pi/3, and 2pi

Answer 2

First of all, treat this as any polynomial of the form
2y^2 - 5y - 3 =0, where y = sin(3x)

y = (-5 +/- sqrt(25 + 24)) / 4

y = (-5 +/- 7) / 4
y = -12/4 or 2/4
y = -3 or y = 1/2

sin(3x) = -3 or sin(3x) = 1/2
3x = invsin(-3) or 3x = invsin(1/2)
3x is not defined or 3x = pi/6

x = pi/18

Answer 3

Try factoring the quadratic, with sin(3x) = u:

2u² - 5u - 3 = 0
(2u +1)(u - 3) = 0

Then put sin(3x) back:

2sin(3x) + 1 = 0
sin 3x = -1/2
3x = 7π/6 + 2πn
x = (7/18)π + (2π/3)n
or
3x = -π/6 + 2πn
x = -π/18 + (2π/3)n

so x = 7π/18, 11π/18, 19π/18, 23π/18, 31π/18, 35π/18

Answer 4

call sin3x =z
2z^2-5z-3=0
so
z=((5+-sqrt25+24)/4
z= 3 and z= -1/2
z= 3 cant be because sin <1
so sin 3x=-1/2
3x=7pi/6 +2kpi and 3x = 11pi/6+2kpi
so x= 7pi/18 +2kpi/3
and x= 11pi/18 +2kpi/3 k integer
so we get x= 7pi/18,19pi/18 and 31pi/18
x= 11pi/18 ,23pi/18 and 35pi/18 all six in (0,2pi)

Answer 5

Transform, transform, transform.
If we let sin(3x)= Z, then we have
2Z^2 - 5Z - 3=0
This can be factored, and you will have 2 roots. Only roots of [1] or less count. Then you can figure out where sin(3x)= root occurs on 0,2pi.
Go to it.

Answer 6

Factor:

(2sin3x+1)(sin3x-3)=0

Continue solving ;)