Can you write those two expressions in the form of Csin(x+φ), where C>=0 and -π<=φ<=π?

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = x and tha: b = φ

sin(x + φ) = sin(x).cos(φ) + cos(x).sin(φ) → you multiply by C

C.sin(x + φ) = C.sin(x).cos(φ) + C.cos(x).sin(φ) → let: C.cos(φ) = 2 ← equation (1)

C.sin(x + φ) = 2.sin(x) + C.cos(x).sin(φ) → let: C.sin(φ) = 3 ← equation (2)

C.sin(x + φ) = 2.sin(x) + 3.cos(x) ← this is your expression

You can get a system of 2 equations:

(1) : C.cos(φ) = 2

(2) : C.sin(φ) = 3

You calculate (2)/(1)

[C.sin(φ)]/[C.cos(φ)] = 3/2

sin(φ)/cos(φ) = 3/2

tan(φ) = 3/2

φ ≈ 56.31° → 0.9828 rd

You calculate (1)² + (2)²

[C.cos(φ)]² + [C.sin(φ)]² = 2² + 3²

C².cos²(φ) + C².sin²(φ) = 4 + 9

C².[cos²(φ) + sin²(φ)] = 13 → recall the famous formula: cos²(φ) + sin²(φ) = 1

C² = 13

C = √13

Recall your expression: 2.sin(x) + 3.cos(x) = C.sin(x + φ)

2.sin(x) + 3.cos(x) = √13 * sin(x + 0.9828)

You know now this identity

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = x and tha: b = φ

sin(x + φ) = sin(x).cos(φ) + cos(x).sin(φ) → you multiply by C

C.sin(x + φ) = C.sin(x).cos(φ) + C.cos(x).sin(φ) → let: C.cos(φ) = 7 ← equation (1)

C.sin(x + φ) = 7.sin(x) + C.cos(x).sin(φ) → let: C.sin(φ) = 4 ← equation (2)

C.sin(x + φ) = 7.sin(x) + 4.cos(x) ← this is your expression

You can get a system of 2 equations:

(1) : C.cos(φ) = 7

(2) : C.sin(φ) = 4

You calculate (2)/(1)

[C.sin(φ)]/[C.cos(φ)] = 4/7

sin(φ)/cos(φ) = 4/7

tan(φ) = 4/7

φ ≈ 29.74° → 0.5191 rd

You calculate (1)² + (2)²

[C.cos(φ)]² + [C.sin(φ)]² = 4² + 7²

C².cos²(φ) + C².sin²(φ) = 16 + 49

C².[cos²(φ) + sin²(φ)] = 65 → reclla the famous formula: cos²(φ) + sin²(φ) = 1

C² = 65

C = √65

Recall your expression: 7.sin(x) + 4.cos(x) = C.sin(x + φ)

7.sin(x) + 4.cos(x) = √65 * sin(x + 0.5191)