Given that A is acute and 2sin^2A-1÷3=cos^2A, find angle A

Answer 1

To solve this problem, we can use the trigonometric identity for sin^2(A) + cos^2(A) = 1.

Given that 2sin^2(A) - 1 = 3cos^2(A), we can substitute sin^2(A) = 1 - cos^2(A) into the equation above to get:

2(1 - cos^2(A)) - 1 = 3cos^2(A)
2 - 2cos^2(A) - 1 = 3cos^2(A)
2 - 2cos^2(A) - 1 - 3cos^2(A) = 0
-5cos^2(A) - 1 = 0
5cos^2(A) = -1
cos^2(A) = -1/5

Since A is acute, cos(A) > 0. Let's find the square root of -1/5 to get:

cos(A) = sqrt(-1/5)
cos(A) = sqrt(-1)/sqrt(5)
cos(A) = i.sqrt(1)/sqrt(5)
cos(A) = i/sqrt(5)

Since cos(A) is a real number, there's no acute angle A that satisfies the given equation. Therefore, there is no solution to this problem.