To solve the inequality cos(2x - π/3) > √2/2, we need to first find the values of x that satisfy the inequality.
Using the trigonometric identity cos(2x) = 2cos^2(x) - 1, we can rewrite the inequality as:
2cos^2(x) - 1 - √2/2 > 0
Now, let's solve for cos(x):
2cos^2(x) - 1 - √2/2 > 02cos^2(x) > 1 + √2/2cos^2(x) > (1 + √2/2) / 2cos^2(x) > (2 + √2) / 4
Taking the square root of both sides and considering the restrictions on the range of cos(x), we get:
cos(x) > √((2 + √2) / 4)cos(x) > √(2 + √2) / 2
So, the solution to the inequality is:
x ∈ (arccos(√(2 + √2) / 2) + 2πn, 2π - arccos(√(2 + √2) / 2) + 2πn) for n ∈ Z
To solve the inequality cos(2x - π/3) > √2/2, we need to first find the values of x that satisfy the inequality.
Using the trigonometric identity cos(2x) = 2cos^2(x) - 1, we can rewrite the inequality as:
2cos^2(x) - 1 - √2/2 > 0
Now, let's solve for cos(x):
2cos^2(x) - 1 - √2/2 > 0
2cos^2(x) > 1 + √2/2
cos^2(x) > (1 + √2/2) / 2
cos^2(x) > (2 + √2) / 4
Taking the square root of both sides and considering the restrictions on the range of cos(x), we get:
cos(x) > √((2 + √2) / 4)
cos(x) > √(2 + √2) / 2
So, the solution to the inequality is:
x ∈ (arccos(√(2 + √2) / 2) + 2πn, 2π - arccos(√(2 + √2) / 2) + 2πn) for n ∈ Z