Question

1) cos(x-1)=√3/2 2)cosx>√2/2 , x ∈ ( 0 ; π/2 )

Answer 1

$1) \: \cos(x - 1) = \frac{ \sqrt{3} }{2} \\ \\ a) \: x - 1 = \frac{\pi}{6} + 2\pi \: n \\ x = \frac{\pi}{6} + 1 + 2\pi \: n \\ \\ b) \: x - 1 = - \frac{\pi}{6} + 2\pi \: n \\ x = - \frac{\pi}{6} + 1 + 2\pi \: n$

n € Z

$2) \: \cos(x) > \frac{ \sqrt{2} }{2} \\ \\ - \frac{\pi}{4} + 2\pi \: n < x < \frac{\pi}{4} + 2\pi \: n$
n € Z

Answer 2

1. To solve the equation cos(x - 1) = √3/2, we can use the inverse cosine function or arccosine. Applying the inverse function to both sides of the equation, we have:

x - 1 = arccos(√3/2)

Since the cosine function has a range of [-1, 1], arccos returns values in the interval [0, π]. Thus, the equation becomes:

x - 1 = π/6

To isolate x, we add 1 to both sides:

x = π/6 + 1

So the solution to the equation cos(x - 1) = √3/2 is x = π/6 + 1.

2. We want to solve the inequality cos(x) > √2/2, where x is in the interval (0, π/2).

Since the cosine function is positive in the first quadrant (0, π/2), we can use the inverse cosine function to find the range of values for x that satisfy the inequality.

cos(x) > √2/2 is equivalent to x > arccos(√2/2).

The value of arccos(√2/2) is π/4. However, we need to find values of x greater than π/4 that lie in the interval (0, π/2).

Therefore, the solution to the inequality cos(x) > √2/2, where x is in the interval (0, π/2), is x > π/4.