Question

Sin(3п/2-2x)=-√2 / 2

Answer 1

$\sin\left(\dfrac{3\pi}{2} -2x\right)=-\dfrac{\sqrt{2} }{2}$

$\dfrac{3\pi}{2} -2x=(-1)^{k+1}\dfrac{\pi}{4} +\pi k$

$2x=\dfrac{3\pi}{2} -(-1)^{k+1}\dfrac{\pi}{4} +\pi k$

$x=\dfrac{3\pi}{4} -(-1)^{k+1}\dfrac{\pi}{8} +\dfrac{\pi k}{2}$

$x=\dfrac{3\pi}{4} +(-1)^{k+2}\dfrac{\pi}{8} +\dfrac{\pi k}{2}$

Упростим показатель степени, так как нас интересует только его четность:

$x=\dfrac{3\pi}{4} +(-1)^{k}\dfrac{\pi}{8} +\dfrac{\pi k}{2},\ k\in\mathbb{Z}$

Answer 2

To solve the trigonometric equation sin(3π/2 - 2x) = -√2 / 2, you can follow these steps:

1. Identify the reference angle: The reference angle is the angle between the terminal side of the angle and the x-axis in the standard position. In this case, the reference angle would be π/4 radians because sin(π/4) = √2 / 2.

2. Rewrite the equation: The equation sin(3π/2 - 2x) = -√2 / 2 can be rewritten as sin(π/2 + π - 2x) = -√2 / 2. This is because sin(3π/2) = -1.

3. Apply the sine addition formula: The sine addition formula states that sin(A + B) = sin(A)cos(B) + cos(A)sin(B). In this case, A = π/2, B = -2x.

   sin(π/2 + (-2x)) = sin(π/2)cos(-2x) + cos(π/2)sin(-2x) 1 * cos(-2x) + 0 * sin(-2x) = cos(-2x) cos(-2x) = -√2 / 2

4. Solve for x: Since cos(-2x) = cos(2x) and cos(2x) is positive in the second and third quadrants, we need to find the angles that satisfy this equation in those quadrants.

   Using the unit circle or a calculator, find the angles θ in the second and third quadrants where cos(θ) = -√2 / 2.

   In the second quadrant: θ = 3π/4 In the third quadrant: θ = 5π/4

   Now, set up equations for the angles: 2x = 3π/4 and 2x = 5π/4

   Solve for x in each equation: x = 3π/8 and x = 5π/8

So, the solutions for the equation sin(3π/2 - 2x) = -√2 / 2 are x = 3π/8 and x = 5π/8, as long as x falls in the appropriate domains of those solutions.