solve cosx= (squareroot3)/ 2 where 0 <(with a line underneath) x <(with a line underneath) 2 pie
a) how many solutions are possible?
b) in which quadrants would you find the solutions?
c) determine the related angle for the equation.
d) determine all the solutions for the equation
we know that
cos π/6 = √3/2
and
cos(π±x) = -cos(x)
cos(2π±x) = cos(x)
Steve is correct with the reasoning. Anyways, the answers are:
a) 2
b) I and IV
c) π/6 radians (30 degrees) and -π/6 (330 degrees)
d) 2nπ ± π/6 radians (n=any integer)
To solve the equation cos(x) = √3/2, where 0 < x < 2π:
a) To determine how many solutions are possible, we need to identify the number of times the cosine function intersects the value √3/2 within the given interval of 0 to 2π. The cosine function has a period of 2π, meaning it repeats itself every 2π units. Within one period, it intersects the value √3/2 twice. Therefore, in the given interval, there will be two solutions.
b) To determine in which quadrants the solutions lie, we can refer to the unit circle. The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants. Since we are looking for cos(x) = √3/2, which is positive, the solutions will be in the first and fourth quadrants.
c) To determine the related angle for the equation, we can find the inverse cosine of √3/2. Inverse cosine, or arccosine, is denoted as cos^(-1). Evaluating cos^(-1)(√3/2) gives us π/6 radians or 30 degrees.
d) To find all the solutions, we need to identify the values of x within the given interval where cos(x) = √3/2. Since cos(x) has a period of 2π, we can write the general solutions as x = π/6 + 2πn and x = 11π/6 + 2πn, where n is an integer.
Considering the given interval of 0 < x < 2π, we substitute n = 0:
For the first solution:
x = π/6 + 2π(0)
x = π/6
For the second solution:
x = 11π/6 + 2π(0)
x = 11π/6
Therefore, the solutions for the equation cos(x) = √3/2, where 0 < x < 2π, are x = π/6 and x = 11π/6.