Exercise 72 Page 918

To solve the given equation, we will first factor it and use the Zero Product Property to solve for $sin θ .$ Then we will use the unit circle to find the exact values of $θ$ that satisfy the equation.

Solving the Equation for $sin θ$

Let's start by rewriting the equation so that all of the terms are on the left-hand side, so we can factor it.

2 sin^{2} θ = - sin θ

AddEqn

$LHS + sin θ = RHS + sin θ$

2 sin^{2} θ + sin θ = 0

FactorOut

Factor out $sin θ$

sin θ (2 sin θ + 1) = 0

Next, we can use the Zero Product Property to solve the equation for

sin θ .

sin θ (2 sin θ + 1) = 0

ZeroProdProp

Use the Zero Product Property

sin θ = 0 2 sin θ + 1 = 0 (I) (II)

▼

(II):

Solve for

sin θ

SubEqn

$(II):$ $LHS - 1 = RHS - 1$

sin θ = 0 2 sin θ = - 1

DivEqn

$(II):$ $LHS / 2 = RHS / 2$

sin θ = 0 sin θ = \frac{- 1}{2}

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

sin θ = 0 sin θ = - \frac{1}{2}

Exercise 72 Page 918

Solving the Equation for sinθ

Solving the Equation for $sin θ$