Solve for x given the interval of [0,2pi).
cos^2x=2+2sinx
I got (3pi)/2 or 270 degrees..
By the way, does the notation [0,2pi) mean
0<=x<2 pi or 360?
The reason they usually use [0,2pi) is that 0 and 2pi are really the same angle. So, your solution set need not include them both.
cos^2x-2sinx-2 = 0
1-sin^2x-2sinx-2 = 0
sin^2x+2sinx+1 = 0
sinx = -1
so, yes x = 3pi/2
Ah, solving equations and unraveling the mysteries of notation! You're on the right track, my friend. [0,2pi) means 0 <= x < 2pi or 0 degrees <= x < 360 degrees.
Let's solve the equation now!
cos^2x = 2 + 2sinx
First, rearrange things a bit:
cos^2x - 2sinx - 2 = 0
Now, substituting 1 - sin^2x for cos^2x:
(1 - sin^2x) - 2sinx - 2 = 0
Rearranging again:
-sin^2x - 2sinx - 1 = 0
Now it's time to let our clown math skills shine! Let's use a substitution to make things easier.
Let's substitute sinx with a variable, let's say t. Now our equation is:
-t^2 - 2t - 1 = 0
What a funny little quadratic equation we've got here! To solve it, we can either factor or use the quadratic formula.
Using the quadratic formula, we find:
t = (-(-2) ± sqrt((-2)^2 - 4(-1)(-1))) / (2(-1))
t = (2 ± sqrt(4 - 4)) / (-2)
t = (2 ± sqrt(4)) / (-2)
t = (2 ± 2) / (-2)
So we have two values for t:
t1 = 4 / -2 = -2
t2 = 0 / -2 = 0
Now, we substitute these values back into the original equation sinx = t:
For t1 = -2: sinx = -2
For t2 = 0: sinx = 0
Since the range of sinx is [-1, 1], we discard the value of -2 for sinx.
Lastly, we find the corresponding values of x:
For sinx = 0:
x = 0 degrees or 180 degrees
So, the solutions for x in the interval [0,2pi) are 0 degrees and 180 degrees.
Keep up the good work, my mathematical friend!
To solve the equation cos^2x = 2 + 2sinx over the interval [0, 2pi), you can follow these steps:
Step 1: Rewrite the equation using the identity cos^2x = 1 - sin^2x:
1 - sin^2x = 2 + 2sinx
Step 2: Rearrange the equation to form a quadratic equation:
sin^2x + 2sinx - 3 = 0
Step 3: Factor the quadratic equation:
(sinx - 1)(sinx + 3) = 0
Step 4: Set each factor equal to zero and solve for x by considering the given interval:
For sinx - 1 = 0, add 1 to both sides:
sinx = 1
Considering the interval [0, 2pi), we can find the solutions for sinx = 1 within this interval:
The angle satisfying sinx = 1 is pi/2 or 90 degrees.
For sinx + 3 = 0, subtract 3 from both sides:
sinx = -3
However, there are no solutions to sinx = -3 within the interval [0, 2pi).
Therefore, the only solution that satisfies the equation within the given interval is x = pi/2 or 90 degrees.
Regarding the notation [0,2pi), it represents the interval from 0 to 2pi where 0 is included, and 2pi is excluded. So, it does not include 360 degrees.
To solve the equation cos^2x = 2 + 2sinx, follow these steps:
1. Rearrange the equation to isolate the cosine term: cos^2x - 2cosx - 2sinx = 0.
2. Rewrite the left side of the equation using the identity cos^2x = 1 - sin^2x: 1 - sin^2x - 2cosx - 2sinx = 0.
3. Combine like terms: -sin^2x - 2cosx - 2sinx + 1 = 0.
4. Rearrange the equation: -sin^2x - 2sinx - 2cosx + 1 = 0.
5. Rearrange further: sin^2x + 2sinx + 2cosx - 1 = 0.
6. Use the identity sin^2x + cos^2x = 1 to replace cos^2x with 1 - sin^2x: sin^2x + 2sinx + 2(1 - sin^2x) - 1 = 0.
7. Simplify: sin^2x + 2sinx + 2 - 2sin^2x - 1 = 0.
8. Combine like terms: -sin^2x + 2sinx + 1 = 0.
9. Factor the quadratic: -(sinx - 1)(sinx + 1) = 0.
10. Set each factor equal to zero: sinx - 1 = 0 or sinx + 1 = 0.
11. Solve for x:
- If sinx - 1 = 0, then sinx = 1. This occurs when x = pi/2 or 90 degrees.
- If sinx + 1 = 0, then sinx = -1. This occurs when x = 3pi/2 or 270 degrees.
So, the solutions for x within the interval [0, 2pi) are x = pi/2 or 90 degrees and x = 3pi/2 or 270 degrees.
Regarding the notation [0, 2pi), it means that the values of x in the interval include 0 but not 2pi. In other words, it represents the range of x from 0 to slightly less than 2pi (or 360 degrees), inclusively.