To solve this equation, let's make a substitution to simplify it:
Let y = cos^2(x).
Now we can rewrite the equation as:
16y^2 - 24y + 9 = 0
This is now a quadratic equation that we can solve using the quadratic formula:
y = [ -(-24) ± √((-24)^2 - 4169) ] / 2*16y = [ 24 ± √(576 - 576) ] / 32y = [ 24 ± 0 ] / 32y = 24 / 32y = 3/4
Now substitute back in cos^2(x) for y:
cos^2(x) = 3/4
Taking the square root of both sides, we get:
cos(x) = ±√(3/4)cos(x) = ±√3 / 2
Therefore, the solutions are:
x = ±π/6 and x = ±5π/6.
To solve this equation, let's make a substitution to simplify it:
Let y = cos^2(x).
Now we can rewrite the equation as:
16y^2 - 24y + 9 = 0
This is now a quadratic equation that we can solve using the quadratic formula:
y = [ -(-24) ± √((-24)^2 - 4169) ] / 2*16
y = [ 24 ± √(576 - 576) ] / 32
y = [ 24 ± 0 ] / 32
y = 24 / 32
y = 3/4
Now substitute back in cos^2(x) for y:
cos^2(x) = 3/4
Taking the square root of both sides, we get:
cos(x) = ±√(3/4)
cos(x) = ±√3 / 2
Therefore, the solutions are:
x = ±π/6 and x = ±5π/6.