Solve: cos2x(2cosx+1) for 0degrees is less than or equal to x is equal to or less than 360degrees
To solve the inequality, we need to find the values of x that satisfy the given condition.
Let's simplify the expression first:
cos2x(2cosx+1) = cos^2x - sin^2x + 2cosx + 1
= 1 - sin^2x + 2cosx + 1
= 2cosx - sin^2x + 2
Now, the inequality becomes:
2cosx - sin^2x + 2 ≤ 0
To solve this inequality, we'll break it down into two separate inequalities:
1. Solve 2cosx - sin^2x + 2 = 0 for x:
2cosx - sin^2x + 2 = 0
sin^2x = 2cosx + 2
sin^2x = 2(1 - sin^2x) + 2
sin^2x = 4 - 2sin^2x
3sin^2x = 4
sin^2x = 4/3
sinx = ±√(4/3)
We will focus on sinx = √(4/3) since sinx cannot be greater than 1.
To find the value of x:
x = arcsin(√(4/3))
x ≈ 61.93°
2. Solve 2cosx - sin^2x + 2 < 0 for x:
2cosx - sin^2x + 2 < 0
sin^2x - 2cosx > 2
Now, we need to draw the unit circle and find the values of x where sin^2x - 2cosx > 2. We can also use the identity sin^2x + cos^2x = 1 to simplify cosx in terms of sinx.
sin^2x - 2cosx = 2
sin^2x - 2√(1 - sin^2x) = 2
sin^2x - 2√(1 - sin^2x) - 2 = 0
At this point, we can use numerical methods or a graphing calculator to find approximate solutions for x. A possible solution is:
x ≈ 208.144°
Therefore, the solution to the inequality cos2x(2cosx+1) ≤ 0° for 0° ≤ x ≤ 360° is:
0° ≤ x ≤ 61.93° or 208.144° ≤ x ≤ 360°