Solve the following trigonometric inequality, given that 0 ≤ x ≤ 2π. cos2x > -

So x ε{ π/6 , 5π/6 , 3π/2 }

Step-by-step explanation:

Let's break this down shall we?

If cos (2x) = sin(x) then,

1 − 2 sin^2(x) = sin(x)

2sin^2(x) + sin(x) − 1 = 0

So we must now substitute,

k = sin(x)

2k^2 + k − 1 = 0

(2k−1) (k+1) = 0

sin (x) = 1/2 or sin(x)=−1

If sin (x)=1/2 (for 0 ≤ x ≤ 2π)

x = π/6 = 30° or x = 5π/6 = 150°

If sin (x) = − 1 (for 0 ≤ x ≤ 2π)

x = 3π/2 = 270

So x ε{ π/6 , 5π/6 , 3π/2 }

(or their equivalent in degrees)

answer: i don't understand it

step-by-step explanation: sorry