+0  
 
0
606
1
avatar

Solve the equation on the interval 0 ≤ θ  <  2π

3sin(θ/2) = 3/2

 Oct 7, 2014

Best Answer 

 #1
avatar+130511 
+5

3sin(θ/2) = 3/2      divide both sides by 3

sin(θ/2) = 1/2

Now the sine θ = 1/2 when θ =  pi/6 and 5pi/6

So, set θ/2 =pi/6

Multiply both sides by 2

θ = pi/3

And set θ/2 = 5pi/6

Multiply both sides by 2

θ = 10pi/6 = 5pi/3

So...our two answers are pi/3 and 5pi/3 in the given interval

 

 Oct 7, 2014
 #1
avatar+130511 
+5
Best Answer

3sin(θ/2) = 3/2      divide both sides by 3

sin(θ/2) = 1/2

Now the sine θ = 1/2 when θ =  pi/6 and 5pi/6

So, set θ/2 =pi/6

Multiply both sides by 2

θ = pi/3

And set θ/2 = 5pi/6

Multiply both sides by 2

θ = 10pi/6 = 5pi/3

So...our two answers are pi/3 and 5pi/3 in the given interval

 

CPhill Oct 7, 2014

0 Online Users