Find all real numbers that satisfy the equation.
2 cos x + 1 = 0
2cosx = -1
cosx = -1/2
I know that cos 60° = cos π/6 = + 1/2
and the cosine is negative in quads II or III
x = 180-60 = 120° or x = 180+60 = 240°
or in radians
x = 2π/3 or x = 4π/3
the cosx function has a period of 360° or 2π radians, so
adding or subtracting 360° to any degree answer or adding/subtracting 2π to any radian answer will produce a new answer.
To find all real numbers that satisfy the equation 2 cos x + 1 = 0, we need to solve for x.
Step 1: Subtract 1 from both sides of the equation:
2 cos x = -1
Step 2: Divide both sides of the equation by 2:
cos x = -1/2
Now, we need to find the values of x that make the cosine of x equal to -1/2. To do this, we can examine the unit circle or the values of cosine on the interval [0, 2π].
On the unit circle, the cosine of an angle represents the x-coordinate of the point where the terminal side of the angle intersects the unit circle.
The cosine of an angle is -1/2 at two different angles on the unit circle: π/3 and 5π/3. These two angles give us the solutions for our equation.
On the interval [0, 2π], we can find the solutions by looking at the values of cosine for each angle. Starting from 0 radians, we can find that cos(π/3) = 1/2 and cos(5π/3) = -1/2.
Therefore, the solutions to the equation 2 cos x + 1 = 0 are:
x = π/3 + 2πn, where n is an integer
x = 5π/3 + 2πn, where n is an integer
These equations give us all the possible values of x that satisfy the equation 2 cos x + 1 = 0.