How many solutions does the equation cosx + 1/2 = 1 have for 0<x<2pi
cosx+1/2=1
cosx+1=2
cosx=2-1
cosx=1
therefore 1 solution?
A) 1
b) 2
c) 3
d) 4
What did you do ????
cosx + 1/2 = 1
cosx = 1/2
the cosine is positive in quads I and IV
x = 30° or x = 330° or (π/6 and 11π/6)
so obviously 2 solutions between 0 and 2π
How come I am not suppose to get rid of the the denominator?
if you had multiplied by 2 then your second line should have been
2cosx + 1 = 2 . and then
2cosx = 1
cosx = 1/2 , now pick it up in my solution.
you forgot to multiply the cosx by 2
Well, let's examine the situation closely, with a pinch of humor, of course!
So, we have the equation cos(x) + 1/2 = 1. Now, for all you math enthusiasts out there, it's important to remember that cosine is a sly little function, always playing hide and seek with us. In this case, cos(x) loves to prance around in the range of -1 to 1. But what happens when it joins forces with 1/2 in a quest to be equal to 1?
Now, if we subtract 1/2 from both sides, we get cos(x) = 1/2. Ah, delicious! We've successfully trapped cosine!
Now, thinking back to our trigonometry adventures, we know that cosine equals 1/2 at two points in the interval 0 < x < 2π: once at x = π/3 and again at x = 5π/3. These values of x make up our magical, mystical, and hilarious solutions.
So, the answer is B) 2 solutions. Two brave warriors, ready to prove that cosine and 1/2 can indeed tango together.
To find the number of solutions to the equation cosx + 1/2 = 1 for 0 < x < 2pi, we need to solve the equation and determine how many solutions satisfy the given conditions.
First, let's solve the equation:
cosx + 1/2 = 1
Subtract 1/2 from both sides:
cosx = 1 - 1/2
cosx = 1/2
Now, to find the solutions of this equation for 0 < x < 2pi, we need to consider the unit circle and the cosine function's values in this range.
In the unit circle, the cosine function is positive (greater than zero) in the first and fourth quadrants. The cosine function equals 1/2 in the first and fourth quadrants.
In the first quadrant, the value of x is between 0 and pi/2, and in the fourth quadrant, the value of x is between 3pi/2 and 2pi.
So, the solutions for x that satisfy cosx = 1/2 and 0 < x < 2pi are:
x = pi/3 (in the first quadrant)
x = 5pi/3 (in the fourth quadrant)
Therefore, there are 2 solutions for the given equation when 0 < x < 2pi.
The correct answer is (B) 2.