Find the solution.
1. sqrt2 sin x + 1 = 0
My ans: x = 5pi/4 + 2npi and 7pi/4 + 2npi
2. sec x - 2 = 0
My ans: x = pi/3 + 2npi and 5pi/3 + 2npi
To find the solution to the given equations, we can follow these steps:
1. sqrt(2) sin(x) + 1 = 0:
- Subtract 1 from both sides: sqrt(2) sin(x) = -1
- Divide both sides by sqrt(2): sin(x) = -1/sqrt(2)
- To find the solutions, we need to find the angles whose sine is equal to -1/sqrt(2). Since -1/sqrt(2) is a special value, we can reference the unit circle to determine the angles.
- On the unit circle, the sine value of -1/sqrt(2) occurs at two particular angles: -3π/4 and 3π/4.
- Since sin(x) is a periodic function, we need to consider all possible values of x. The general solution for sin(x) = -1/sqrt(2) is x = -3π/4 + 2nπ and x = 3π/4 + 2nπ, where n is an integer.
- Simplifying the general solution:
- For the first set of angles: x = -3π/4 + 2nπ
- x = -3π/4, -3π/4 + 2π, -3π/4 + 4π, -3π/4 + 6π, ...
- For the second set of angles: x = 3π/4 + 2nπ
- x = 3π/4, 3π/4 + 2π, 3π/4 + 4π, 3π/4 + 6π, ...
- So, the solution to the equation sqrt(2) sin(x) + 1 = 0 is x = -3π/4 + 2nπ and x = 3π/4 + 2nπ, where n is an integer.
2. sec(x) - 2 = 0:
- Add 2 to both sides: sec(x) = 2
- To find the solutions, we need to find the angles whose secant is equal to 2. The secant function is the reciprocal of the cosine function.
- On the unit circle, the cosine value of 1/2 occurs at two particular angles: π/3 and 5π/3.
- Since sec(x) is a periodic function, we need to consider all possible values of x. The general solution for sec(x) = 2 is x = π/3 + 2nπ and x = 5π/3 + 2nπ, where n is an integer.
- Simplifying the general solution:
- For the first set of angles: x = π/3 + 2nπ
- x = π/3, π/3 + 2π, π/3 + 4π, π/3 + 6π, ...
- For the second set of angles: x = 5π/3 + 2nπ
- x = 5π/3, 5π/3 + 2π, 5π/3 + 4π, 5π/3 + 6π, ...
- So, the solution to the equation sec(x) - 2 = 0 is x = π/3 + 2nπ and x = 5π/3 + 2nπ, where n is an integer.