solve for all 0≤x≤2π

Question

1. √2 csc²x+csc x=√2
2. sec x+cos x=tan x sin x-1

Answer 1

1) Let's start with the first one: √2 csc²x + csc x = √2

We can let csc x = y, so we get an equation that looks like this: √2 y² + y - √2 = 0

We can solve this for y = csc x quite simply. Just use the quadratic formula:

y = - 1 ± √(1² - 4(√2)(-√2) / 2√2

y = -1 ± √ 1+ 4(2) /2√2

y = -1 ± 3 / (2√2)

so y = -1 - 3 / 2√2 or y = -1 + 3 / 2√2

y = -4 / 2√2 or y = 2 / 2√2

y = -2/√2 or y = 1/√2

y = -√2 or y = 1 /√2

Since y = csc x = 1/ sin x, we have either that 1/ sin x = -√2 or 1/cos x = 1/√2

So in this case we have either sin x = -√2/2 or sin x = √2. Since | sin x |≤ 1 for all x, only the first equation has solutions, which are x = -π/4 or x = -3π/4, neither of which are in your domain, so the equation has no solutions on the interval 0 ≤ x ≤ 2π.

2) Now onto the second equation: sec x + cos x = tan x sin x -1

We can write this in a more convinient way: 1/ cos x + cos x = (sin x/ cos x) *sin x -1

1 / cos x + cos x = sin²x / cos x - 1

1 / cos x - sin²x / cos x = -cos x - 1

Since we have a sin²x and we know that sin²x = 1 - cos²x, we can now write this entire equation in terms of cos x only. Here's what we get:

1/ cos x - (1 - cos²x) / cos x = -cos x - 1

cos²x / cos x = -cos x - 1

cos x = - cos x - 1

2 cos x = -1

cos x = -1/2

This has a solution at x = 2π /3.