Question

2*16^cosx -9*4^cosx +4=0

Answer 1

4^cosx=t. 2t^2-9t+4=0. d=81-4*2*4=49. t1=9-7/2*4=1/4. t2=9+7/2*4=2. cosx=1/2=+-p/3+2pk

Answer 2

To solve the equation 2 * 16^cos(x) - 9 * 4^cos(x) + 4 = 0, we can simplify it by using a substitution. Let's substitute y = 4^cos(x):

2 * (4^2)^cos(x) - 9 * (4^cos(x)) + 4 = 0

Simplifying further:

2 * 4^(2 * cos(x)) - 9 * (4^cos(x)) + 4 = 0

Now substitute y = 4^cos(x) back into the equation:

2 * y^2 - 9 * y + 4 = 0

This is now a quadratic equation in terms of y. We can solve it using the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = -9, and c = 4. Substituting these values into the quadratic formula:

y = (-(-9) ± √((-9)^2 - 4 * 2 * 4)) / (2 * 2) y = (9 ± √(81 - 32)) / 4 y = (9 ± √49) / 4

Taking the square root:

y = (9 ± 7) / 4

We have two possible solutions for y:

1. y = (9 + 7) / 4 = 16 / 4 = 4
2. y = (9 - 7) / 4 = 2 / 4 = 1/2

Now we substitute the values of y back into the equation y = 4^cos(x):

1. 4 = 4^cos(x)

Taking the logarithm base 4 of both sides:

log4(4) = log4(4^cos(x)) 1 = cos(x)

2. 1/2 = 4^cos(x)

Taking the logarithm base 4 of both sides:

log4(1/2) = log4(4^cos(x)) log4(1) - log4(2) = cos(x) 0 - log4(2) = cos(x)

Therefore, the solutions to the equation 2 * 16^cos(x) - 9 * 4^cos(x) + 4 = 0 are:

1. cos(x) = 1
2. cos(x) = -log4(2)