Solve the equation 2cos(x)+1=0,0≤x≤2π0≤x≤2π. Show all of your work.
you have cosx = -1/2,
your reference angle is π/3
That means there is a solution in QII and QII, where cosx is negative
...
would it be 120 degrees
To solve the equation 2cos(x) + 1 = 0, we can start by isolating the cosine term.
Step 1: Subtract 1 from both sides of the equation:
2cos(x) = -1
Step 2: Divide both sides of the equation by 2:
cos(x) = -1/2
Now, we need to find all the values of x within the given range (0 ≤ x ≤ 2π) that satisfy the equation cos(x) = -1/2.
To determine these values, we can refer to the unit circle or the cosine function graph.
On the unit circle, we know that cosine represents the x-coordinate of a point on the circle. The unit circle shows that cos(x) = -1/2 occurs at π/3 and 5π/3 (in the second and third quadrants).
Therefore, we have two values of x that satisfy the equation: x = π/3 and x = 5π/3.
However, we must ensure that these values fall within the given range, 0 ≤ x ≤ 2π.
For x = π/3, it falls within the range since 0 ≤ π/3 ≤ 2π.
For x = 5π/3, we notice that it is greater than 2π but not greater than 4π (which is the next full revolution). Therefore, we can subtract 2π from 5π/3 to bring it within the given range:
x = 5π/3 - 2π = π/3.
So, the values of x that satisfy the equation 2cos(x) + 1 = 0 within the given range are x = π/3 and x = 5π/3.