Question

2(cosx-1)sin2x=3cos(3п/2+x)[3п/2;3п]​

Answer 1

$2(\cos{x}-1)2\sin{x}\cos{x}-3\cos{(3\pi/2+x)}=0\\\cos{(3\pi/2+x)}=\sin{x}\\4\sin{x}(\cos^2{x}-\cos{x}-3/4)=0\\D=1+3=2^2\\\sin{x}(\cos{x}+1/2)(\cos{x}-3/2)=0\\!|\cos{x}|\leq 1!\\\left[\begin{array}{ccc}\sin{x}=0\\\cos{x}=-1/2\\\end{array}$

x={pi*n;±2pi/3+2pi*n}, n∈Z.

Ответ: x={2pi; 8pi/3; 3pi}

Answer 2

To solve the trigonometric equation 2(cosx-1)sin2x = 3cos(3π/2+x) over the interval [3π/2, 3π], follow these steps:

Step 1: Simplify the right-hand side of the equation. Step 2: Expand the left-hand side of the equation. Step 3: Solve for x.

Step 1: Simplify the right-hand side of the equation: cos(3π/2 + x) = cos(3π/2)cos(x) - sin(3π/2)sin(x) Since cos(3π/2) = 0 and sin(3π/2) = -1: cos(3π/2 + x) = -sin(x)

Step 2: Expand the left-hand side of the equation: 2(cosx - 1)sin2x = 2(cosx - 1) * 2sin(x)cos(x) = 4sin(x)cos(x)(cosx - 1)

Step 3: Substitute the simplified right-hand side back into the equation: 4sin(x)cos(x)(cosx - 1) = 3(-sin(x))

Now, we can cancel out the common factor of sin(x): 4cos(x)(cosx - 1) = -3

Expand the expression: 4cos^2(x) - 4cos(x) = -3

Rearrange the equation to bring everything to one side: 4cos^2(x) - 4cos(x) + 3 = 0

Now, this is a quadratic equation in terms of cos(x). To solve it, we can use the quadratic formula:

cos(x) = [ -(-4) ± √((-4)^2 - 4 * 4 * 3) ] / (2 * 4) cos(x) = [ 4 ± √(16 - 48) ] / 8 cos(x) = [ 4 ± √(-32) ] / 8 cos(x) = [ 4 ± 4i√2 ] / 8

cos(x) = (1/2) ± (1/2)i√2

Since the domain of cos(x) is [-1, 1], the imaginary part is not possible in this context. Therefore, we have only one real solution for cos(x):

cos(x) = (1/2)

Now, to find the value of x, we take the arccosine (inverse cosine) of (1/2):

x = arccos(1/2)

Using the unit circle or trigonometric identities, we know that arccos(1/2) is π/3.

However, we need to ensure that the solution lies in the given interval [3π/2, 3π].

π/3 is approximately 1.0472, which lies between π/2 and π. It doesn't fall within the interval [3π/2, 3π].

To find the other solution within the given interval, we consider that cos(x) is positive in both the first and fourth quadrants (0 ≤ x ≤ π/2 and 2π ≥ x ≥ 3π/2).

So, the other solution within the given interval is:

x = 2π - π/3 = 6π/3 - π/3 = 5π/3

Thus, the solutions to the equation within the interval [3π/2, 3π] are:

x = π/3 and x = 5π/3.