Answer:
a) 4 cosx^4 - cos²2x=0
We can start by splitting the equation into two parts:
4 cosx^4 = cos²2x
cosx^4 = (1/4)cos²2x
From here, we can use the identity:
cos²2x = (cos 2x)² = (2cos²x -1)²
So, we have:
cosx^4 = (1/4)(2cos²x -1)²
Now we can square root both sides:
cosx^2 = (1/2) (2cos²x -1)
Now we square both sides again:
cos²x = (1/4) (2cos²x -1)
We can now subtract (1/4) from both sides and add cos²x to both sides:
(3/4)cos²x = (1/4)
We can now divide both sides by (3/4) to find the value of cosx:
cosx = sqrt(1/3)
So the solutions of the equation are x = (2n+1)π/3, where n is an integer.
b) sec x tan x =sqrt(2)
We know that sec x = 1/cosx and tan x = sinx/cosx.
So we can substitute these values into the equation:
1/cosx * sinx/cosx = sqrt(2)
This simplifies to:
sinx = sqrt(2)cosx
We can now divide both sides by cosx:
tan x = sqrt(2)
So the solutions of the equation are x = pi/4 + nπ, where n is an integer.
Note that the solutions are restricted to the domain where cos(x) is not equal to 0.
It is important to note that this solution is only a small part of the full solution set and that there might be more solutions in specific intervals.