Question

SOS! SOS! SOS! cos(3x+pi/2)=✓2/2

Answer 1

$Cos(3x+\frac{\pi }{2})=\frac{\sqrt{2} }{2}\\\\-Sin3x=\frac{\sqrt{2} }{2}\\\\Sin3x=-\frac{\sqrt{2} }{2}\\\\3x=(-1)^{n}arcSin(-\frac{\sqrt{2} }{2})+\pi n,n\in Z\\\\3x=(-1)^{n+1}\frac{\pi }{4}+\pi n,n\in Z\\\\\boxed{x=(-1)^{n+1}\frac{\pi }{12}+\frac{\pi n }{3},n\in Z}$

Answer 2

It seems like you're looking for the solutions to the trigonometric equation:

cos(3x + π/2) = √2/2

To find the values of x that satisfy this equation, you can use the properties of the cosine function.

1. First, recall that cos(π/4) = √2/2. So, your equation can be rewritten as:

cos(3x + π/2) = cos(π/4)

2. To find the solutions for 3x + π/2, you can use the fact that the cosine function repeats its values every 2π radians. Therefore, you can write:

3x + π/2 = 2nπ ± π/4, where n is an integer.

3. Now, solve for x:

3x = 2nπ ± π/4 - π/2

3x = 2nπ ± (2π - π/4)/2

3x = 2nπ ± (8π - π)/8

3x = 2nπ ± (7π/8)

x = (2nπ ± 7π/8)/3

So, the general solutions for the equation are:

x = (2nπ ± 7π/8)/3

where n is an integer. These solutions cover all possible values of x that satisfy the given equation.