Part b is same as part a,
Except for part a u let y=0,
Part b, let y = 3,
Bring over the 3 and solve again normally as part (a)

You'll need to use the R-formula for this :

3 = 5sin 2x + 2cos² x

2 + 1 = 5sin 2x + 2cos² x

2 + cos² x + sin² x = 5sin 2x + 2cos² x

2 = 5sin 2x + 2cos² x - cos² x - sin² x

2 = 5sin 2x + cos² x - sin² x

2 = 5sin 2x + cos 2x

Next,


R = √(5² + 1²) = √26

tan α = 1/5

α = tan-¹ (1/5)

So ,


2 = √26 sin (2x + α)

sin (2x + α) = 2/√26

sin (2x + α) = √2 / √13

2x + α = sin-¹ (√2 / √13)

2x = sin-¹ (√2 / √13) - α

Basic angle for 2x + α ≈ 23.0935°
(Just key in sin-¹ (√2 / √13) into calculator)

Acute angle for α ≈ 11.3099°
(Just key in tan-¹ (1/5) into calculator)


Since √2 / √13 is positive, we look at the 1st and 2nd quadrant .


Since 0° < x < 360°, then 0° < 2x < 720°

2x ≈ 23.0935° - 11.3099°, 180° - 23.0935° - 11.3099°, 360° + 23.0935° - 11.3099°, 360° + 180° - 23.0935° - 11.3099°

2x ≈ 11.7836°, 145.5966° , 371.7836°, 505.5966°,

x ≈ 5.8918°, 72.7983°, 185.8918°, 252.7983°

x ≈ 5.89°, 72.80° , 185.89°, 252.80° (2d.p)

Update : alternative method


3 = 5sin 2x + 2cos² x

3(sin² x + cos² x) = 5(2sin x cos x) + 2cos² x

3sin² x + 3 cos² x = 10 sin x cos x + 2cos² x

3sin² x / cos² x + 3 = 10sin x / cos x + 2

3tan² x + 3 = 10tan x + 2

3tan² x - 10tan x + 1 = 0


Note that this is in the quadratic form 3X² - 10X + 1 = 0, where X = tan x

Using the Quadratic formula ,

tan x = (-(-10)±√(10² - 4(3)(1)) ) / (2(3))

tan x = (10±√88) / 6

tan x = (10±2√22) / 6

tan x = (5±√22) / 3

(both are positive so look at the first and third quadrant)

Basic angle = tan-¹ (5-√22) / 3 ≈ 5.892°

x ≈ 5.892°, 5.892° + 180°
x ≈ 5.892°, 185.892°
x ≈ 5.89°, 185.9° (1d.p)

Or

Basic angle = tan-¹ (5+√22) / 3 ≈ 72.798°

x ≈ 72.798°, 72.798° + 180°

x ≈ 72.798° , 252.798°

x ≈ 72.8°, 252.8° (1d.p)