solve for all 0≤x≤2π

Question

1. tan⁴x-2=tan²x+sec²x
2. cos x-cot x=0

Answer 1

1) tan⁴(x) - 2 = tan²(x) + sec²(x)

Using the Pythagorean Identity 1 + tan²(x) = sec²(x) substitute to get:

tan⁴(x) - 2 = tan²(x) + 1 + tan²(x)

tan⁴(x) - 2 = 2tan²(x) + 1

tan⁴(x) - 2tan²(x) - 3 = 0

Let w = tan²(x) so the equation becomes:

w² - 2w - 3 = 0

(w - 3)(w + 1) = 0

w = 3 and w = -1

Back-substituting tan²(x) for w we get:

tan²(x) = 3 or tan(x) = ±√3 and tan²(x) = -1 or tan(x) = √-1 (not a real number solution)

So for tan(x) = ±√3, using the unit circle, we get x = π/3, x = 2π/3, x = 4π/3, x = 5π/3

2) cos(x) - cot(x) = 0

cos(x) - cos(x)/sin(x) = 0

cos(x)sin(x)/sin(x) - cos(x)/sin(x) = 0

(cos(x)sin(x) - cos(x))/sin(x) = 0

The denominator cannot contribute to the left side of the equation being equal to zero, therefore, we can just look at the numerator:

cos(x)sin(x) - cos(x) = 0

cos(x)[sin(x) - 1] = 0

cos(x) = 0 so x = π/2 and 3π/2

sin(x) - 1 = 0

sin(x) = 1 so x = π/2

So x = π/2 and x = 3π/2