Question

The number of xϵ[0,2π] for which |√2sin⁴x+18cos²x−√2cos⁴x+18sin²x|=1 is:

a. 4
b. 2
c. 6
d. 8

Answer 1

There are no values of x in the interval [0, 2π] for which the equation |√2sin⁴x+18cos²x−√2cos⁴x+18sin²x|=1 is true.

Step-by-step explanation:

To solve the equation |√2sin⁴x+18cos²x−√2cos⁴x+18sin²x|=1, we can simplify it by using trigonometric identities. First, notice that √2sin⁴x+18cos²x−√2cos⁴x+18sin²x can be written as (√2sin²x + 18cos²x)(√2sin²x + 18sin²x).

Using the identity sin²x + cos²x = 1, this expression becomes (1 + 16cos²x)(1 + 16sin²x). We now have the equation (1 + 16cos²x)(1 + 16sin²x) = 1. Expanding this equation gives us 1 + 16cos²x + 16sin²x + 256cos²xsin²x = 1. Simplifying further, we get 16(cos²x + sin²x) + 256cos²xsin²x = 0.

Since cos²x + sin²x = 1, this equation simplifies to 16 + 256cos²xsin²x = 0.

Now, let's focus on the equation 256cos²xsin²x = -16. Dividing both sides by 16 gives us 16cos²xsin²x = -1. Taking the square root of both sides gives us 4cosxsinx = ±sqrt(-1). Notice that the square root of -1 is an imaginary number, so there are no solutions for this equation.

Therefore, there are no values of x in the interval [0, 2π] for which the equation |√2sin⁴x+18cos²x−√2cos⁴x+18sin²x|=1 is true.