Find the solution set of each equation if 0° ≤ θ < 360°. (Please explain it with proper solutions)

Question

15 (sin 2θ)^2 + sin 2θ − 2 = 0

Answer 1

Make a simple substitution to make this problem easier. Let w = sin2θ. Then re-writing the problem gives you:

15w² + w - 2 = 0

Factoring:

(5w + 2)(3w - 1) = 0

Setting each binomial equal to zero gives:

5w + 2 = 0 or w = -2/5

3w - 1 = 0 or w = 1/3

Now, back substitute sin2θ = w to get:

sin2θ = -2/5 and sin2θ = 1/3

Use inverse sine:

2θ = sin^-1(-2/5) and 2θ = sin^-1(1/3)

Use you calculator to find the FIRST angle where sine is -2/5. I got -23.578° (a Q4 answer). Convert this into a standard 0 - 360° measurement by adding 360° to it giving 336.422°. However, sine is negative also in Q3. To get that answer, consider the Unit Circle. You will also get -2/5 for an angle 180° + 23.578° = 203.578°. You can get the same sine value for 203.578° + 360° = 563.578° and for 336.422° + 360° = 696.422°

Remember that this answer is actually 2θ so:

2θ = 336.422° means θ = 168.21°

2θ = 203.578° means θ = 101.79°

2θ = 563.578° means θ = 281.79°

2θ = 696.422° means θ = 348.21°

Now do the same with 2θ = sin^-1(1/3)

2θ = 19.471° (a Q1 answer)

Using the unit circle, the Q2 answer is 2θ = 180 - 19.471 = 160.529°

You also get the same answer for 19.471° + 360° = 379.471°

You also get the same answer for 160.529° + 360° = 520.529°

2θ = 19.471° means θ = 9.74°

2θ = 160.529° means θ = 80.26°

2θ = 379.471° means θ = 189.74°

2θ = 520.529° means θ = 260.26°

So you get a total of 8 answers.

Answer 2

sin(2θ) = (- 1 ± √121)/30 = (- 1 ± 11)/30; sin(2θ) = 1/3 or sin(2θ) = - 2/5;

2θ = sin^-1(1/3) + 360°n; θ = 9.74° + 180°n; θ = 9.74°, 189.74°

2θ = 180° - 19.74° + 360°n = 160.26° + 360°n; θ = 80.13° + 180°n; θ = 80.13°, 160.13°.

2θ = 180° + sin^-1(2/5) + 360°n; θ = 78.21° + 180°n, θ = 101.79, 281.79°

2θ = 360° - sin^-1(2/5) + 360°n; θ = 168.21°, 348.21°