2sin2x = cos2x for - pi x pi

Answer 1

To solve the equation 2sin(2x) = cos(2x) for -π ≤ x ≤ π, we can use the double-angle formula sin(2x) = 2sin(x)cos(x) and rewrite the equation as:

2(2sin(x)cos(x)) = cos(2x)

Expanding the equation further:

4sin(x)cos(x) = cos(2x)

Using the double-angle formula again for cosine, cos(2x) = cos^2(x) - sin^2(x), we have:

4sin(x)cos(x) = cos^2(x) - sin^2(x)

Rearranging the equation:

4sin(x)cos(x) + sin^2(x) = cos^2(x)

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can substitute it into the equation:

4sin(x)cos(x) + sin^2(x) = 1 - sin^2(x)

Simplifying:

4sin(x)cos(x) + 2sin^2(x) - 1 = 0

Now we have a quadratic equation in terms of sin(x). Let's substitute y = sin(x):

4ycos(x) + 2y^2 - 1 = 0

Multiplying through by 2:

8y^2 + 4ycos(x) - 2 = 0

Using the quadratic formula, y = [-4cos(x) ± √(16cos^2(x) - 4(8)(-2))] / (2(8))

Simplifying:

y = [-4cos(x) ± √(64cos^2(x) + 64)] / 16

y = [-cos(x) ± √(cos^2(x) + 1)] / 2

Now substitute y back as sin(x):

sin(x) = [-cos(x) ± √(cos^2(x) + 1)] / 2

We can solve for sin(x) by considering the two cases of + and -:

1) sin(x) = [-cos(x) + √(cos^2(x) + 1)] / 2

Rearranging:

cos(x) = -[1 + 2sin(x)] / √(4sin^2(x) + 1)

Since -π ≤ x ≤ π, we can use the unit circle to determine where the cosine values are negative or positive. From the unit circle, we can see that the cosine in the given range is positive at x = -π/2 and x = π/2.

For x = -π/2:

cos(-π/2) = -[1 + 2sin(-π/2)] / √(4sin^2(-π/2) + 1)

0 = -1 / √(4 + 1)

This is not true, so x = -π/2 is not a valid solution.

For x = π/2:

cos(π/2) = -[1 + 2sin(π/2)] / √(4sin^2(π/2) + 1)

0 = -1 / √(4 + 1)

This is not true, so x = π/2 is not a valid solution.

Therefore, the equation 2sin(2x) = cos(2x) has no solutions in the range -π ≤ x ≤ π.