Question

2 sin ( 3п/2 + 2x) = 1

Answer 1

$2sin( \frac{3 \pi }{2}+2x)=1 \\ sin( \frac{3 \pi }{2}+2x)= \frac{1}{2} \\ \frac{3 \pi }{2}+2x=(-1)^k* \frac{ \pi }{6} + \pi k \\ 2x=(-1)^k^+^1* \frac{4 \pi }{3}+ \pi k \\ x=(-1)^k^+^1* \frac{2 \pi }{3} + \frac{ \pi }{2}k$
x=(-1)^k+1*2pi/3+pi/2*k, k∈Z

Answer 2

To solve the equation 2 sin(3π/2 + 2x) = 1, we can follow these steps:

Step 1: Simplify the equation Since sin(3π/2 + 2x) = sin(3π/2)cos(2x) + cos(3π/2)sin(2x) = -cos(2x), we can rewrite the equation as: 2(-cos(2x)) = 1

Step 2: Divide both sides by -2 Dividing both sides of the equation by -2, we get: cos(2x) = -1/2

Step 3: Solve for 2x To find the solutions for cos(2x) = -1/2, we need to find the angles whose cosine is -1/2. One such angle is π/3. Since cosine is an even function, another solution is the reflection of π/3 about the y-axis, which is 2π/3.

So, we have: 2x = π/3 + 2kπ or 2x = 2π/3 + 2kπ

where k is an integer.

Step 4: Solve for x Dividing both sides of each equation by 2, we find: x = π/6 + kπ or x = π/3 + kπ

where k is an integer.

Therefore, the solutions to the equation 2 sin(3π/2 + 2x) = 1 are: x = π/6 + kπ or x = π/3 + kπ

where k is an integer.