Chapter 1.

1.1 Problem Statement

Solve for 0 ≤ θ ≤ 2π.

sin(θ) = sin(2θ)

1.2 Step 1

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

Question Sequence

Question 1.1

sin(2θ) = 0a5Zo3IErDwQ3l14IGs+fYA138CaIdOnBLQJ63rqt/hA1T/xSzVNPowkLDHtXDuWb/zWYmt9MYKUMUBG4LZtndr0JaSCOagKnleGTg==

Incorrect

Correct

1.3 Step 2

Substitute sin(2θ) = 2sin(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 1.2

sin(θ) - 2sin(θ)cos(θ) = 1Wh3cvJ2xF4=

Incorrect

Correct

Question 1.3

sin(θ)(1-2cos(θ)) = 0

Thus, we must find θ where either sin(θ) or 1-2·cos(θ) = 1Wh3cvJ2xF4=

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

1.4 Step 3

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

Question Sequence

Question 1.4

θ = 1Wh3cvJ2xF4= (smaller value)

θ = π (larger value)

Correct.

Incorrect.

1.5 Step 4

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

1 − 2cos(θ) = 0

cos(θ) =

Question 1.5

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

θ = 96riwYEHPfT6J+0o5bJE9Dzxn6XXishO

θ = iDX3J1Zo1csEQl8nllBo/A5Ip1+Fmj/lZeErIQ==

Correct.

Incorrect.

1.6 Step 5

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

Question Sequence

Question 1.6

iZMKYAeKFaO9e4+8ft8yTFNuj6qTPQAH8IcrbUuMjFjfFrveHUnGUEEk+XWrVkCkm7mu7bDP8S7zACUrhq5PrxrrSTgC/LbZzhjDyg==

Correct.

Incorrect.