Question

2cos((Pi//4-3X))=2root(2)

Answer 1

2cos(π/4-3x)=√2
cos(π/4-3x)=√2/2
π/4-3x=+/-π/4+2πn, n∈Z
π/4-3x=π/4+2πn            π/4-3x=-π/4+2πk
-3x=2πn                                -3x=-π/2+2πk
x=-2/3πn, n∈Z                          x=π/6-2/3πk, k∈Z

Answer 2

To solve the equation 2cos((π/4 - 3x)) = 2√2, we can start by isolating the cosine term on one side of the equation.

Dividing both sides by 2, we get:

cos((π/4 - 3x)) = √2

Now, to solve for x, we need to find the inverse cosine (also known as arccos) of both sides of the equation. The inverse cosine will cancel out the cosine function, leaving us with:

(π/4 - 3x) = arccos(√2)

To isolate x, we can rearrange the equation:

3x = π/4 - arccos(√2)

Finally, we divide both sides by 3 to solve for x:

x = (1/3) * (π/4 - arccos(√2))

Please note that the exact numerical value of x will depend on the specific value of arccos(√2).