To solve the equation sin(x-π/4) = 1, we first need to find the general solution for sin(x) = 1. This occurs when x = π/2 + 2πn, where n is an integer.
Now, we need to find when x-π/4 = π/2 + 2πn. Adding π/4 to both sides gives us x = π/2 + π/4 + 2πn, which simplifies to x = 3π/4 + 2πn.
Therefore, the solution to the equation sin(x-π/4) = 1 is x = 3π/4 + 2πn, where n is an integer.
To solve the equation 2cos^2(x) - 5cos(x) + 2 = 0, we first need to factor it.
The equation can be factored into (2cos(x) - 1)(cos(x) - 2) = 0.
Setting each factor to zero gives us:
2cos(x) - 1 = 0 cos(x) - 2 = 0
Solving the first equation, we get cos(x) = 1/2. This occurs when x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.
Solving the second equation, we get cos(x) = 2, which has no real solutions.
Therefore, the solutions to the equation 2cos^2(x) - 5cos(x) + 2 = 0 are x = π/3 + 2πn and x = 5π/3 + 2πn, where n is an integer.
Now, we need to find when x-π/4 = π/2 + 2πn. Adding π/4 to both sides gives us x = π/2 + π/4 + 2πn, which simplifies to x = 3π/4 + 2πn.
Therefore, the solution to the equation sin(x-π/4) = 1 is x = 3π/4 + 2πn, where n is an integer.
To solve the equation 2cos^2(x) - 5cos(x) + 2 = 0, we first need to factor it.The equation can be factored into (2cos(x) - 1)(cos(x) - 2) = 0.
Setting each factor to zero gives us:
2cos(x) - 1 = 0
cos(x) - 2 = 0
Solving the first equation, we get cos(x) = 1/2. This occurs when x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.
Solving the second equation, we get cos(x) = 2, which has no real solutions.
Therefore, the solutions to the equation 2cos^2(x) - 5cos(x) + 2 = 0 are x = π/3 + 2πn and x = 5π/3 + 2πn, where n is an integer.