solve for x: 4sin^2x= 4cosx+1
4sin^2 = 4cos + 1
4 - 4cos^2 = 4cos + 1
4cos^2 + 4cos - 3 = 0
(2cos-1)(2cos+3) = 0
so, cosx = 1/2 or -3/2
-3/2 is not a choice, so just find where cosx = 1/2.
answers: A. pi/3, 5pi/3
B. pi/3, 4pi/3
C. pi/3, 2pi/3
D. none
To solve for x in the equation 4sin^2x = 4cosx + 1, we can use trigonometric identities to simplify the equation and then solve for x algebraically.
Step 1: Rewrite sin^2x using the identity sin^2x = 1 - cos^2x.
4(1 - cos^2x) = 4cosx + 1
Step 2: Simplify the equation.
4 - 4cos^2x = 4cosx + 1
Step 3: Rearrange the equation and combine like terms to form a quadratic equation.
4cos^2x + 4cosx - 3 = 0
Step 4: Solve the quadratic equation by factoring, completing the square, or using the quadratic formula.
Since there is no clear factoring method or suitable quadratic formula here, we can use a substitution to simplify the equation further.
Let's substitute cosx = t.
4t^2 + 4t - 3 = 0
Step 5: Solve the quadratic equation for t using factoring, completing the square, or the quadratic formula.
In this case, we can use the quadratic formula.
The quadratic formula is t = (-b ± sqrt(b^2 - 4ac)) / 2a, where for our equation a = 4, b = 4, and c = -3.
t = (-4 ± sqrt(4^2 - 4 * 4 * -3)) / (2 * 4)
t = (-4 ± sqrt(16 + 48)) / 8
t = (-4 ± sqrt(64)) / 8
t = (-4 ± 8) / 8
Simplifying further:
t = 4/8 = 1/2
t = -12/8 = -3/2
Step 6: Solve for x.
Since we substituted t = cosx, we need to convert t back to cosx.
For t = 1/2:
cosx = 1/2
Taking the inverse cosine or arccos of both sides:
x = arccos(1/2)
x = π/3
For t = -3/2:
cosx = -3/2
Taking the inverse cosine or arccos of both sides:
x = arccos(-3/2)
x = undefined (no real solution)
Therefore, the only solution to the given equation is x = π/3.