Question

2sin(3x-π/4)-√2=0 (0°, 90°)

Answer 1

2 * sin (3x - п/4) = -√2; 

sin (3x - п/4) = -√2/2; 

3x - п/4 = (-1)^n * arcsin (-√2/2) + pi * n, n ∈ Z;  

3x - п/4 = (-1)^n * 5 * п/4 + pi * n, n ∈ Z; 

Числа и переменные нужно записать по одну сторону от уравнения. Значения х оставляем на левой части уравнения, а все остальное перенесем на противоположную сторону. 

3x  = (-1)^n * 5 * п/4 + п/4 + pi * n, n ∈ Z;  

x  = (-1)^n * 5 * п/12 + п/12 + pi/3 * n, n ∈ Z.  

Answer 2

To solve the equation 2sin(3x-π/4) - √2 = 0 in the interval (0°, 90°), we can use algebraic manipulation and trigonometric identities. Let's solve it step by step:

1. Start with the given equation: 2sin(3x-π/4) - √2 = 0.

2. Add √2 to both sides to isolate the sine term: 2sin(3x-π/4) = √2.

3. Divide both sides by 2: sin(3x-π/4) = √2/2.

4. Since we are looking for solutions in the interval (0°, 90°), we need to find the reference angle for √2/2.

5. The reference angle for √2/2 is π/4 radians or 45°.

6. Now, we need to find the angles within the interval (0°, 90°) that have a sine value of √2/2. These angles are 45° and 90° - 45° = 45°.

7. Set up two equations based on the angles:

   • 3x - π/4 = π/4 (45°) and
   • 3x - π/4 = π - π/4 (90° - 45°).

8. Solve the first equation: 3x - π/4 = π/4. 3x = π/4 + π/4. 3x = π/2. x = π/6.

9. Solve the second equation: 3x - π/4 = π - π/4. 3x = π - π/4 + π/4. 3x = π. x = π/3.

10. The solutions for x in the given equation within the interval (0°, 90°) are x = π/6 and x = π/3.

Therefore, the solutions to the equation 2sin(3x-π/4) - √2 = 0 in the interval (0°, 90°) are x = π/6 and x = π/3.