Answer:
We are given the equation: cos(x) + sqrt(3) = -cos(x)
Combine Cos(x) terms:
Add cos(x) to both sides to isolate it on the left:
2cos(x) + sqrt(3) = 0
Isolate sqrt(3):
Subtract sqrt(3) from both sides:
2cos(x) = -sqrt(3)
Divide by the coefficient of cos(x):
Important Note: Dividing by cosine (especially when it's negative) can lead to extraneous solutions. We'll address this later.
cos(x) = -sqrt(3) / 2
Find the inverse cosine (arccos):
Take the arccos of both sides to find the values of x where cosine equals -sqrt(3) / 2. Keep in mind that the arccos function only outputs values in the range [0, pi].
arccos(cos(x)) = arccos(-sqrt(3) / 2)
x = arccos(-sqrt(3) / 2)
Finding all solutions in [0, 2pi):
Cosine is negative in both Quadrant II and Quadrant III. Since arccos only outputs values in [0, pi], we need to consider the periodicity of cosine to find all solutions within the given interval.
Solution in Quadrant II:
The arccos function gives us one solution, x = (5π/6). Because cosine has a period of 2π, adding multiples of 2π will also be solutions:
x = (5π/6) + 2πn where n is any integer
Solution in Quadrant III:
Since cosine is negative in Quadrant III as well, we can find the mirror image of the solution in Quadrant II across the y-axis (pi). This gives us another solution:
x = pi + (5π/6) = (11π/6)
Similar to Quadrant II, adding multiples of 2π will also be solutions here:
x = (11π/6) + 2πn where n is any integer
Addressing Extraneous Solutions:
Remember, we divided by cosine earlier. While cos(x) = -sqrt(3) / 2 has solutions at (5π/6) and (11π/6), these solutions only hold true if cosine is actually defined at those points.
Cosine is undefined at pi/2 + (2πn) for any integer n. This eliminates solutions where x = (5π/6) + (2πn) because it translates to x = (pi/2) + (4πn/3), which falls within these undefined points.
Similarly, cosine is undefined at (3π/2) + (2πn) for any integer n. This eliminates solutions where x = (11π/6) + (2πn) because it translates to x = (3π/2) + (8πn/3), which also falls within undefined points.
Therefore, the only solutions within the interval [0, 2pi) are:
x = 5π/6
x = 11π/6
