4*cos(t)^2<1 inequation

Given the inequality:
$$4 \cos^{2}{\left(t \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$4 \cos^{2}{\left(t \right)} = 1$$
Solve:
Given the equation
$$4 \cos^{2}{\left(t \right)} = 1$$
transform
$$4 \cos^{2}{\left(t \right)} - 1 = 0$$
$$4 \cos^{2}{\left(t \right)} - 1 = 0$$
Do replacement
$$w = \cos{\left(t \right)}$$
This equation is of the form

a*w^2 + b*w + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$w_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$w_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 4$$
$$b = 0$$
$$c = -1$$
, then

D = b^2 - 4 * a * c = 

(0)^2 - 4 * (4) * (-1) = 16

Because D > 0, then the equation has two roots.

w1 = (-b + sqrt(D)) / (2*a)

w2 = (-b - sqrt(D)) / (2*a)

or
$$w_{1} = \frac{1}{2}$$
$$w_{2} = - \frac{1}{2}$$
do backward replacement
$$\cos{\left(t \right)} = w$$
Given the equation
$$\cos{\left(t \right)} = w$$
- this is the simplest trigonometric equation
This equation is transformed to
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
Or
$$t = \pi n + \operatorname{acos}{\left(w \right)}$$
$$t = \pi n + \operatorname{acos}{\left(w \right)} - \pi$$
, where n - is a integer
substitute w:
$$t_{1} = \pi n + \operatorname{acos}{\left(w_{1} \right)}$$
$$t_{1} = \pi n + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$t_{1} = \pi n + \frac{\pi}{3}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(w_{2} \right)}$$
$$t_{2} = \pi n + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$t_{2} = \pi n + \frac{2 \pi}{3}$$
$$t_{3} = \pi n + \operatorname{acos}{\left(w_{1} \right)} - \pi$$
$$t_{3} = \pi n - \pi + \operatorname{acos}{\left(\frac{1}{2} \right)}$$
$$t_{3} = \pi n - \frac{2 \pi}{3}$$
$$t_{4} = \pi n + \operatorname{acos}{\left(w_{2} \right)} - \pi$$
$$t_{4} = \pi n - \pi + \operatorname{acos}{\left(- \frac{1}{2} \right)}$$
$$t_{4} = \pi n - \frac{\pi}{3}$$
$$t_{1} = \frac{\pi}{3}$$
$$t_{2} = \frac{2 \pi}{3}$$
$$t_{3} = \frac{4 \pi}{3}$$
$$t_{4} = \frac{5 \pi}{3}$$
$$t_{1} = \frac{\pi}{3}$$
$$t_{2} = \frac{2 \pi}{3}$$
$$t_{3} = \frac{4 \pi}{3}$$
$$t_{4} = \frac{5 \pi}{3}$$
This roots
$$t_{1} = \frac{\pi}{3}$$
$$t_{2} = \frac{2 \pi}{3}$$
$$t_{3} = \frac{4 \pi}{3}$$
$$t_{4} = \frac{5 \pi}{3}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$t_{0} < t_{1}$$
For example, let's take the point
$$t_{0} = t_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
=
$$- \frac{1}{10} + \frac{\pi}{3}$$
substitute to the expression
$$4 \cos^{2}{\left(t \right)} < 1$$
$$4 \cos^{2}{\left(- \frac{1}{10} + \frac{\pi}{3} \right)} < 1$$

     2/1    pi\    
4*sin |-- + --| < 1
      \10   6 /    

but

     2/1    pi\    
4*sin |-- + --| > 1
      \10   6 /    

Then
$$t < \frac{\pi}{3}$$
no execute
one of the solutions of our inequality is:
$$t > \frac{\pi}{3} \wedge t < \frac{2 \pi}{3}$$

         _____           _____  
        /     \         /     \  
-------ο-------ο-------ο-------ο-------
       t1      t2      t3      t4

Other solutions will get with the changeover to the next point
etc.
The answer:
$$t > \frac{\pi}{3} \wedge t < \frac{2 \pi}{3}$$
$$t > \frac{4 \pi}{3} \wedge t < \frac{5 \pi}{3}$$