Question

16cos^4 (x) + sin^2 (x) =7/4

Answer 1

16cos^4 x + 1-cos^2 x= 7/4
Пусть cos^2 x = t
Тогда 16t^2 - t- 0,75=0
D= 1+48=49
T1=1+7/32=0,25
T2=1-7/32=-0,1875
1)cos^2x=0,25
Cosx=+- 1/2
2) cos^2x=0,1875

Answer 2

To solve the equation 16cos^4(x) + sin^2(x) = 7/4, we can make use of the trigonometric identity cos^2(x) + sin^2(x) = 1. Let's express the equation in terms of cos(x) only:

16cos^4(x) + sin^2(x) = 7/4

Using the identity sin^2(x) = 1 - cos^2(x), we substitute it into the equation:

16cos^4(x) + 1 - cos^2(x) = 7/4

Rearranging the terms, we get:

16cos^4(x) - cos^2(x) + 1 - 7/4 = 0

Multiplying through by 4 to clear the fraction:

64cos^4(x) - 4cos^2(x) + 4 - 7 = 0

Simplifying further:

64cos^4(x) - 4cos^2(x) - 3 = 0

Now, let's substitute a new variable u = cos^2(x):

64u^2 - 4u - 3 = 0

This is a quadratic equation in u. We can solve it using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / (2a)

Applying the values a = 64, b = -4, and c = -3:

u = (-(-4) ± √((-4)^2 - 4 * 64 * (-3))) / (2 * 64) u = (4 ± √(16 + 768)) / 128 u = (4 ± √784) / 128 u = (4 ± 28) / 128

This gives us two possible values for u:

1. u = (4 + 28) / 128 = 32 / 128 = 1/4
2. u = (4 - 28) / 128 = -24 / 128 = -3/16

Since cos^2(x) cannot be negative, we discard the second solution, u = -3/16.

Now, we solve for cos(x) using the first solution, u = 1/4:

cos^2(x) = 1/4

Taking the square root of both sides:

cos(x) = ±√(1/4)

cos(x) = ±1/2

This gives us two possible values for cos(x):

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

Finally, we can find the corresponding values of x by taking the inverse cosine (arccos) of each solution:

1. x = arccos(1/2)
2. x = arccos(-1/2)

These are the solutions to the equation 16cos^4(x) + sin^2(x) = 7/4. Note that there may be multiple solutions for x within a certain range, typically between 0 and 2π, depending on the context or requirements of the problem.