Chapter 10 - NCPN

Solution
Add 1 to both sides of the equation and then divide both sides by 4.
4sin 2 θ – 1 + 1 = 0 + 1
4sin 2 θ = 1
sin 2 θ = 1 4
(sin θ) 2 = 1 4
Find the square root of both sides of the equation.
sin θ =
1
4
sin θ = ± 1 2
Use the inverse sine function and a graphing calculator to solve for θ.
sin –1 (sin θ) = sin −1 1
2
θ ≈ 0.52
Because the sine function is also positive in the second quadrant,
π – 0.52 ≈ 2.62 radians is also a solution to the trigonometric equation. The
values 0.52 + π ≈ 3.66 and 2.62 + π ≈ 5.76 are also solutions to the equation.
So the solutions for 0 ≤ θ ≤ 2π are 0.52, 2.62, 3.66, and 5.76 radians.
Factoring
Ongoing Assessment
Solve –27cos 2 θ + 3 = 0 for 0 ≤ θ < 2π.
Example 3
Solving by Factoring
Solve 4cos θ sin θ + 2sin θ = 0 for 0 ≤ θ < 2π.
Solution
Divide both sides of the equation by 2 and then factor sin θ from each term.
4cos θ sin θ + 2sin θ = 0
2cos θ sin θ + sin θ = 0
sin θ (2cos θ + 1) = 0
486 Chapter 10 Trigonometric Functions and Identities