Problem Statement

Solve for 0 ≤ θ ≤ 2π.

sin(θ) = sin(2θ)

 
Step 1

Notice that the given equation has a double angle. Recall the double-angle formula for the sine function.

Question Sequence

Question 1

sin(2θ) =

Incorrect
Correct

 
Step 2

Substitute sin(2θ) = 2·sin(θ)cos(θ) into sin(θ) = sin(2θ) and algebraically rearrange the equation so we have a product of two factors set equal to zero.

sin(θ) = sin(2θ)

sin(θ) = 2·sin(θ)cos(θ)

Question Sequence

Question 2

sin(θ) - 2·sin(θ)cos(θ) =

Incorrect
Correct

Question 3

sin(θ)(1-2·cos(θ)) = 0

Thus, we must find θ where either sin(θ) or 1-2·cos(θ) =

Note that we did not divide by sin(θ) at any of the steps because we could be dividing by zero since there are values of θ where 0 ≤ θ < 2π and sin(θ) = 0.

Incorrect
Correct

 
Step 3

For 0 ≤ θ < 2π, there are two values of θ where sin(θ) = 0.

Question Sequence

Question 4

θ = (smaller value)

θ = π (larger value)

Correct.
Incorrect.

 
Step 4

Solve 1 − 2·cos(θ) = 0 for cos(θ) and then solve for θ.

1 − 2·cos(θ) = 0

cos(θ) =

Question 5

For 0 ≤ θ < 2π, there are two values of θ where cos(θ) = .

θ =

θ =

Correct.
Incorrect.

 
Step 5

List all the values of θ that satisfy sin(θ) = sin(2θ) and 0 ≤ θ < 2π.

Question Sequence

Question 6

A.
B.
C.
D.
Correct.
Incorrect.