Could you please show me how to solve this question.If 2 cos A-1=0, where 0< or = A < 2pi what is A equal to?
To solve the equation 2cos(A) - 1 = 0, where 0 ≤ A < 2π, we can follow these steps:
Step 1: Add 1 to both sides of the equation:
2cos(A) = 1
Step 2: Divide both sides by 2:
cos(A) = 1/2
Now, to find the value of A, we need to determine at which angles the cosine function equals 1/2. This can be done by analyzing the unit circle or using the inverse cosine function (also known as arccosine or cos^(-1)). Since we are given that A falls between 0 and 2π, we can use the inverse cosine function to find the solutions within that range.
Step 3: Take the inverse cosine (arccosine) of both sides:
A = arccos(1/2)
The inverse cosine of 1/2 is known to be π/3 (or 60 degrees). However, cosine is positive in both the first and fourth quadrants, so we need to consider the other solution.
Step 4: Since cosine has a periodicity of 2π, and the reference angle for 1/2 is π/3, we can find the other solution by subtracting π/3 from 2π:
A = 2π - π/3
Now we simplify:
A = 6π/3 - π/3
A = 5π/3
Therefore, the solutions for A, where 0 ≤ A < 2π, are A = π/3 and A = 5π/3.