Question

find the general solution of 2sinx=sin3x​

Answer 1

Answer:

on l

Step-by-step explanation:

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

sin(x + x) = sin(x).cos(x) + cos(x).sin(x)

sin(2x) = 2sin(x).cos(x) ← memorize this result as (1)

You know this identity:

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = 2x and suppose that: b = x

sin(2x + x) = sin(2x).cos(x) + cos(2x).sin(x)

sin(3x) = sin(2x).cos(x) + cos(2x).sin(x) ← memorize this result as (2)

Do you know this identity:

cos(a + b) = cos(a).cos(b) - sin(a).sin(b) → suppose that: a = b = x

cos(x + x) = cos(x).cos(x) - sin(x).sin(x)

cos(2x) = cos²(x) - sin²(x) → recall: cos² + sin² = 1 → sin² = 1 - cos²

cos(2x) = cos²(x) - [1 - cos²(x)]

cos(2x) = cos²(x) - 1 + cos²(x)

cos(2x) = 2cos²(x) - 1 ← memorize this result as (3)

2sin(x).sin(3x) = 1 → recall (2)

2sin(x).[sin(2x).cos(x) + cos(2x).sin(x)] = 1 → recall (1)

2sin(x).[2sin(x).cos(x).cos(x) + cos(2x).sin(x)] = 1

2sin(x).[2sin(x).cos²(x) + cos(2x).sin(x)] = 1 → recall (3)

2sin(x).[2sin(x).cos²(x) + {2cos²(x) - 1}.sin(x)] = 1

2sin(x).[2sin(x).cos²(x) + 2cos²(x).sin(x) - sin(x)] = 1

2sin(x).[4sin(x).cos²(x) - sin(x)] = 1

2sin²(x).[4cos²(x) - 1] = 1

2sin²(x).[4cos²(x) - 1] = cos²(x) + sin²(x)

2sin²(x).[4cos²(x) - 1] - cos²(x) - sin²(x) = 0

8sin²(x).cos²(x) - 2sin²(x) - cos²(x) - sin²(x) = 0

8sin²(x).cos²(x) - 3sin²(x) - cos²(x) = 0

8[1 - cos²(x)].cos²(x) - 3[1 - cos²(x)] - cos²(x) = 0

8.cos²(x) - 8cos^4(x) - 3 + 3cos²(x) - cos²(x) = 0

- 8cos^4(x) - 3 + 10cos²(x) = 0

8cos^4(x) - 10cos²(x) + 3 = 0 → let: X = cos²(x) → where: 0 ≤ X ≤ 1

8X² - 10X + 3 = 0

8X² - (4X + 6X) + 3 = 0

8X² - 4X - 6X + 3 = 0

(8X² - 4X) - (6X - 3) = 0

4X(2X - 1) - 3(2X - 1) = 0

(2X - 1)(4X - 3) = 0

First case: (2X - 1) = 0 → 2X = 1 → X = 1/2 → cos²(x) = 1/2

Second case: (4X - 3) = 0 → 4X = 3 → X = 3/4 → cos²(x) = 3/4

First possibility: cos²(x) = 1/2 → cos(x) = ± 1/√2 → cos(x) = ± (√2)/2

When: cos(x) = (√2)/2

x = π/4

x = 2π - (π/4) = 7π/4

When: cos(x) = - (√2)/2

x = π - (π/4) = 3π/4

x = π + (π/4) = 5π/4

Second possibility: cos²(x) = 3/4 → cos(x) = ± (√3)/2

When: cos(x) = (√3)/2

x = π/6

x = 2π - (π/6) = 11π/6

When: cos(x) = - (√3)/2

x = π - (π/6) = 5π/6

x = π + (π/6) = 7π/6

→ Solution = { π/4 ; 3π/4 ; 5π/4 ; 7π/4 ; π/6 ; 5π/6 ; 7π/6 ; 11π/6 }