Given that Ais acute and 2sin^2 A -1/3 A, find angle A.

Answer 1

To find angle A, we need to solve the equation 2sin^2 A - 1/3 = A.

Let's first rearrange the equation to get it in the form of a quadratic equation in terms of sin A:

2sin^2 A - 1/3 = A
2sin^2 A = A + 1/3
sin^2 A = (A + 1/3) / 2
sin A = ±√((A + 1/3) / 2)
(sin A)^2 = (A + 1/3) / 2

Since A is acute, sin A is positive. Therefore:

sin A = √((A + 1/3) / 2)

Squaring both sides gives us:

sin^2 A = (A + 1/3) / 2
(A + 1/3) / 2 = sin^2 A

Now, we know that sin^2 A = 1 - cos^2 A (from trigonometric identity), so we can write:

(A + 1/3) / 2 = 1 - cos^2 A
A + 1/3 = 2 - 2cos^2 A
cos^2 A = 1 - (A + 1/3)/2
cos A = ±√(1 - (A + 1/3)/2)
cos A = ±√((2 - A - 1/3)/2)
cos A = ±√((5/3 - A)/2)

Since A is acute, cos A is positive, so:

cos A = √((5/3 - A)/2)

Now, using the fact that sin^2 A + cos^2 A = 1, we can square and sum the equations for sin A and cos A:

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

Since the sum of the squares of sin A and cos A equals 1, we have a valid solution for angle A.

Therefore, angle A satisfies the equation 2sin^2 A - 1/3 = A.