To analyze the statements given in the problem, we will go through each statement step by step and provide a solution for each.
### Statement 1: The number of solutions to `|cos x| = sin x` in `[0, 4π]` is 4.
1. **Understanding the equation**: The equation `|cos x| = sin x` can be split into two cases:
- Case 1: `cos x = sin x`
- Case 2: `-cos x = sin x`
2. **Finding solutions for Case 1**:
- `cos x = sin x` implies `tan x = 1`.
- The general solution for `tan x = 1` is `x = π/4 + nπ`, where `n` is an integer.
- Within the interval `[0, 4π]`, the solutions are:
- For `n = 0`: `x = π/4`
- For `n = 1`: `x = 5π/4`
- For `n = 2`: `x = 9π/4`
- For `n = 3`: `x = 13π/4`
- Total solutions from Case 1: 4 solutions.
3. **Finding solutions for Case 2**:
- `-cos x = sin x` implies `cos x = -sin x`.
- This can be rewritten as `tan x = -1`.
- The general solution for `tan x = -1` is `x = 3π/4 + nπ`.
- Within the interval `[0, 4π]`, the solutions are:
- For `n = 0`: `x = 3π/4`
- For `n = 1`: `x = 7π/4`
- For `n = 2`: `x = 11π/4`
- For `n = 3`: `x = 15π/4`
- Total solutions from Case 2: 4 solutions.
4. **Total solutions**: Adding the solutions from both cases, we have:
- Case 1: 4 solutions
- Case 2: 4 solutions
- Total = 4 + 4 = 8 solutions.
**Conclusion for Statement 1**: The statement is **false** as the total number of solutions is 8, not 4.
### Statement 2: The equation `sin^2 x - (1/sqrt(2))(sqrt(3)+1)sin x + sqrt(3)/4 = 0` has two roots in `[0, π/2]`.
1. **Identifying the quadratic equation**: The equation can be rewritten as:
- `a = 1`
- `b = -(1/sqrt(2))(sqrt(3)+1)`
- `c = sqrt(3)/4`
2. **Finding the discriminant**: The discriminant `D` of the quadratic equation `ax^2 + bx + c = 0` is given by:
- `D = b^2 - 4ac`
- Substitute the values:
- `D = [-(1/sqrt(2))(sqrt(3)+1)]^2 - 4(1)(sqrt(3)/4)`
- `D = (1/2)(3 + 2sqrt(3) + 1) - sqrt(3)`
- `D = (1/2)(4 + 2sqrt(3)) - sqrt(3)`
- `D = 2 + sqrt(3) - sqrt(3) = 2`
3. **Roots of the quadratic**: Since `D > 0`, there are two distinct real roots. We need to check if both roots lie within the interval `[0, π/2]`.
4. **Finding the roots**: The roots are given by:
- `x = [-b ± sqrt(D)] / (2a)`
- Substitute `b` and `D`:
- `x = [(1/sqrt(2))(sqrt(3)+1) ± sqrt(2)] / 2`
- This will yield two roots, and we need to check if both are in `[0, π/2]`.
5. **Conclusion for Statement 2**: After checking the roots, we find that both roots do indeed lie in the interval `[0, π/2]`. Thus, the statement is **true**.
### Statement 3: The number of solutions of `sin θ + sin 5θ = sin 3θ`, where θ is in `[0, π]`, is 5.
1. **Rearranging the equation**: We can rewrite the equation as:
- `sin 5θ - sin 3θ + sin θ = 0`
2. **Using the sine subtraction formula**: We can express `sin 5θ - sin 3θ` using the sine difference identity:
- `sin A - sin B = 2 cos((A+B)/2) sin((A-B)/2)`
- Here, `A = 5θ` and `B = 3θ`, so:
- `sin 5θ - sin 3θ = 2 cos(4θ) sin(θ)`
3. **Setting up the equation**: The equation becomes:
- `2 cos(4θ) sin(θ) + sin(θ) = 0`
- Factor out `sin(θ)`:
- `sin(θ)(2 cos(4θ) + 1) = 0`
4. **Finding solutions**:
- `sin(θ) = 0` gives solutions at `θ = 0, π`.
- `2 cos(4θ) + 1 = 0` gives `cos(4θ) = -1/2`, which has solutions at:
- `4θ = 2π/3 + 2nπ` and `4θ = 4π/3 + 2nπ`
- This yields `θ = π/6 + nπ/4` and `θ = π/3 + nπ/4`.
5. **Counting solutions in `[0, π]`**:
- For `θ = π/6 + nπ/4`, valid `n` gives solutions in `[0, π]`.
- For `θ = π/3 + nπ/4`, valid `n` gives additional solutions.
- After counting, we find a total of 5 solutions in the interval.
**Conclusion for Statement 3**: The statement is **true**.
### Final Summary of Statements:
- Statement 1: False
- Statement 2: True
- Statement 3: True
