Question

Cos(4x)+4sin^2 (x)=1+2sin^2 (2x)

Answer 1

Cos(4x)+4sin²(x)=1+2sin²(2x)
cos2(2x) +4 ( 1-cos2x)-1-2sin² (2x)=0
                        2
cos²(2x)-sin²(2x)-2sin²(2x)+2(1-cos2x)-1=0
cos²(2x)-3sin²(2x)+2-2cos(2x)-1=0
cos²(2x)-3(1-cos²(2x))-2cos(2x)+1=0
cos²(2x)-3+3cos²(2x)-2cos(2x)+1=0
4cos²(2x)-2cos(2x)-2=0
2cos²(2x)-cos(2x)-1=0
Пусть cos(2x)=y
2y²-y-1=0
D=1+8=9
y₁= 1-3= -2/4=-1/2
       4
y₂= 1+3=1
        4

При у=-1/2
cos(2x)=-1/2
2x= + 2π +2πn, n∈Z
         3
x= +  π + πn, n∈Z
       3

При у=1
cos(2x)=1
2x=2πn, n∈Z
x=πn, n∈Z

Ответ: х= +  π + πn, n∈Z
                  3
           x=πn, n∈Z

Answer 2

We can simplify the left-hand side of the equation using the double angle formula for sine:

sin^2(x) = (1 - cos(2x))/2 4sin^2(x) = 2(1 - cos(2x)) cos(4x) = 2cos^2(2x) - 1

Substituting these expressions into the original equation, we get:

2cos^2(2x) - 1 + 2(1 - cos(2x)) = 1 + 2sin^2(2x)

Simplifying and rearranging terms, we get:

2cos^2(2x) - 2cos(2x) + 1 = 2sin^2(2x)

Using the identity cos^2(x) + sin^2(x) = 1, we can express sin^2(2x) in terms of cos^2(2x):

sin^2(2x) = 1 - cos^2(2x)

Substituting this into the equation, we get:

2cos^2(2x) - 2cos(2x) + 1 = 2(1 - cos^2(2x))

Expanding and simplifying, we get:

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

This is a quadratic equation in cos(2x), which can be solved using the quadratic formula:

cos(2x) = (2 ± sqrt(8))/8

Simplifying, we get:

cos(2x) = (1 ± sqrt(2))/4

Taking the inverse cosine of both sides, we get:

2x = ±cos^-1((1 ± sqrt(2))/4) + 2πn (where n is an integer)

Solving for x, we get:

x = ±(cos^-1((1 ± sqrt(2))/4) + 2πn)/2

Therefore, the solutions to the equation are given by:

x = (cos^-1((1 + sqrt(2))/4) + 2πn)/2 or x = (cos^-1((1 - sqrt(2))/4) + 2πn)/2 (where n is an integer)