Question

Cos3x+2cosx>=0

Answer 1

Cos3x=4cos³x-3cosx
cos3x+2cosx≥0
4cos³x-3cosx +2cosx≥0
4cos³x-cosx≥0
cosx=a
4a³-a≥0
a(2a-1)(2a+1)≥0
a=0  a=1/2  a=-1/2
           _                +                   _               +
----------------[-1/2]----------[0]-----------[1/2]-------------
-1/2≤a≤0 U a≥1/2
-1/2≤cosx≤0 U cosx≥1/2⇒x∈(-∞;∞)

Answer 2

To solve this inequality, we can begin by factoring out the common factor of cos(x) from both terms on the left-hand side:

cos(x)(3cos^2(x) + 2) ≥ 0

Now we have a product of two factors, and the inequality holds true when either both factors are positive or both factors are negative.

First, let's consider the case where both factors are positive:

cos(x) > 0 (since 3cos^2(x) + 2 > 0 for all x)

This is true when x is in the intervals (0, π/2) and (2π, 5π/2).

Next, let's consider the case where both factors are negative:

cos(x) < 0 (since 3cos^2(x) + 2 < 0 for all x)

This is true when x is in the intervals (π/2, 2π) and (5π/2, 4π).

Therefore, the solutions to the inequality cos(3x) + 2cos(x) ≥ 0 are:

x ∈ (0, π/2) ∪ (2π, 5π/2) ∪ (π, 3π/2) ∪ (5π/2, 4π)

or in interval notation:

[0, π/2) ∪ (2π, 5π/2) ∪ (π, 3π/2) ∪ (5π/2, 4π]