Question

Cos 3x = √3/2 [-П;П]

Answer 1

$= \frac{ \cos(3x) \times \sqrt{3} }{2} = \frac{ \sqrt{3} \times \cos(3x) }{2}$

Answer 2

To solve the equation cos(3x) = √3/2 in the interval [-π, π], we need to find the values of x that satisfy this equation.

First, let's find the general solutions for the equation cos(3x) = √3/2:

1. cos(π/6) = √3/2
2. cos(11π/6) = √3/2

Now, we need to find the values of x in the interval [-π, π] that satisfy the equation. The general solutions for cos(3x) = √3/2 are:

1. 3x = π/6
2. 3x = 11π/6

Now, let's solve for x:

1. 3x = π/6 x = π/18

2. 3x = 11π/6 x = 11π/18

However, we need to check if these solutions lie within the interval [-π, π].

For x = π/18: -π < π/18 < π This solution is within the interval.

For x = 11π/18: -π < 11π/18 < π This solution is also within the interval.

So, the solutions for the equation cos(3x) = √3/2 in the interval [-π, π] are: x = π/18 and x = 11π/18.