solve equation sin 2 theta = square root of 3/2?

Answer 1

Let @ = theta.

sin 2@ = sqrt(3) / 2

The sine function equals sqrt(3) / 2 at the following angles measured in degrees:
....,-300, 60, 420, 780,1140,....,60 + 360k and
....,-240, 120, 480, 840, 1200,....,120 + 360k where k is an integer.

2@ = 60 + 360K and
2@ = 120 + 360K
divide both sides of each equation by 2

ANSWER:
in degrees
@ = 30 + 180K and
@ = 60 + 180K where K is an integer.

in radians
@ = pi / 6 + (pi)K and
@ = pi / 3 + (pi)K where K is an integer.

Answer 2

Is it Sin 2theta = sqrt(3)/2 or sqrt (3/2)

If it's the first one, sin(-1) sqrt(3)/2 = pi/3
Since sin is positive in 1st and 2nd quadrant
Then 2theta = pi/3, pi-pi/3, and 2pi + each one...
then divide the answers by 2 such that
theta = pi/6, pi/3, 13pi/6, and so on.

In the other case
if sin 2theta = sqrt(3/2)
You might want to square both sides
such that
sin^2 2theta = 3/2
Then use the identity cos 4theta= 1 - 2sin^2 2theta
You should end up with
cos 4theta = 1-2(3/2) = -2
Since -1<=cos x<=1
This one has no answer - so a long way for no answer.

Answer 3

Hello

The new period is pi or 180. Now there are two cycles in the new graph so when you out in 30 you get the value at 60 and when you put in 60 you get the value at 120 and and at 120 we get 240 and at 150 we get 300 all of which have a sin of sqrt3/2.

So we have : 30, 60, 120, 150

Hope This Helps!!

Answer 4

Will use Ø instead of theta for the simple reason that I can never find theta!
sin 2Ø = √3/2
2Ø = (60 + 360 k)° , 2Ø = (120 + 360 k)°
k = 1,2,3,4,5--------
Ø = (30 + 180 k)° , Ø = (60 + 180 k)°

Answer 5

sin(60°) = √3/2, so if 60° = 2θ, then θ = 30°, or π/6 in radians.

To put it formally:
sin(2θ) = √3/2
2θ = sin^-1 (√3/2)
2θ = 60
θ = 30

Answer 6

Answer= 30

Answer 7

sin2theta=square root of 3/2=sin60degree
2thea=60degree
theta=60degree/2
So, theta=30degree