Use an Addition or Subtraction Formula to simplify the equation.

Question

sin(3θ) cos(θ) − cos(3θ) sin(θ) = 0

 

A.) sin(2ø)=0

 

B.) Find all solutions in the interval [0, 2π).

 

I just need help on part B. Thank you in advance!

Answer 1

Ok, I agree your original equation simplifies to sin(2θ)=0.

 

So now lets take the inverse sin of both sides to solve,

 

sin^-1(sin(2θ))=sin^-1(0)

2θ=sin^-1(0)

 

So you can either plug that into a calculator or just remember (from the unit circle) that the sine is equal to zero at 0 AND at π

 

2θ=0,π

 

But we're not done because the sine is periodic which means there will be solutions every trip around the unit circle (2π)

 

So 2θ=0+N2π and π+N2π where N=0,+/-1,+/-2,etc...

 

But we're looking for values of θ so we have to divide everything by 2

 

θ=0+Nπ and π/2+Nπ 

 

Now we just have to find the values of N that yield solutions on the interval [0,2π]

 

N=0 gives 0 and π/2

N=1 gives π and π3/2

 

So the full solution set is

 

θ=0,π/2,π,π3/2;

 

Good Luck

 

Cheers

 

-Chris

Answer 2

let θ=x

sin(2x)=0 in the interval [0,2∏)

look on the unit circle

sin is 0 at 0º

2x=0º

x=0º

sin is 0 at 180º

2x=180º

x=90º

x=180º because 2*180=360 which brings you back to 0º

2x=180+2∏k for k=0,-1,1,-2,2,...

x=90+∏k for...

x=90+180=270º

x=0º, 90º, 180º, 270º but not 360º because of [0,∏)   (the parenthesis around ∏, otherwise if it were ] around ∏,then 360º would be another answer)