question about calculus?

sin(4x) = cos(4x)

tan(4x) = 1

4x = π/4 + kπ

x = π/16(4k + 1) k = -1, -2 .... -π/2 ≤ x ≤ 0

Answer: -3π/16 and -7π/16

Let a = 4x. The domain of x implies that the domain of a is between -2pi and 0.

Now, there are two places between 0 and -2pi where sin(a)=cos(a)... When a = -3pi/4, sin(a) = cos(a) = -sqrt(2) / 2. When a = pi/4 (which is the same as a = -7pi/4, since we're dealing with negative numbers in our domain), sin(a) = cos(a) = sqrt(2) / 2.

So we want 4x = -3pi/4 and 4x = -7pi/4. This evaluates to x = -3pi/16 and x = -7pi/16. (Both of these fit in our original domain of -pi/3 and 0).

Hope that helps.

sin 4x = cos 4x for -π / 2 ≤ x ≤ 0

sin 4x / cos 4x = 1, for -2π ≤ 4x ≤ 0

tan 4x = 1

The related angle α (acute) satisfies tan α = 1, and hence α = 45° = π / 4 radians. From the ASTC rule, solutions will only occur in quadrants one and three.

So, on -2π ≤ 4x ≤ 0 we have 4x = -2π + α and -π + α

4x = -7π / 4 and -3π / 4

x = -7π / 16 and -3π / 16

On -7π / 16 < x < -3π / 16, sin 4x > cos 4x (I used x = -π / 4 as a test value), and so you know which curve is the 'top' curve and which is the 'bottom'. Thus, the area enclosed by y = sin 4x and y = cos 4x on the interval -π / 2 < x < 0 is given by:

Area = ∫ (from -7π/16 to -3π/16) sin 4x - cos 4x dx = √2 / 2 sq. units