Giải phương trình

sin^2x + cos^2 2x = 1

Assuming the given equation is

sin²(x) + cos²(2x) = 1

Recall the half-angle identities:

sin²(x) = (1 - cos(2x))/2

cos²(x) = (1 + cos(2x))/2

Then the equation can be rewritten as

(1 - cos(2x))/2 + (1 + cos(4x))/2 = 1

Simplifying a bit gives

- cos(2x)/2 + cos(4x)/2 = 0

cos(4x) = cos(2x)

Recall the double angle identity for cosine:

cos(2x) = 2 cos²(x) - 1

Then

cos(4x) = 2 cos²(2x) - 1

and we're left with

2 cos²(2x) - 1 = cos(2x)

or

2 cos²(2x) - cos(2x) - 1 = 0

Factorize the left side and solve for cos(2x) :

(2 cos(2x) + 1) (cos(2x) - 1) = 0

2 cos(2x) + 1 = 0   …   or   …   cos(2x) - 1 = 0

cos(2x) = -1/2   …   or   …   cos(2x) = 1

Solve for x :

[2x = arccos(-1/2) + 2nπ   or   2x = π + arccos(-1/2) + 2nπ]

… or … [2x = 2nπ]

(where n is any integer)

[2x = 2π/3 + 2nπ   or   2x = 4π/3 + 2nπ]

… or … [2x = 2nπ]

[x = π/3 + nπ   or   x = 2π/3 + nπ]   …   or   …   [x = nπ]

the answer is 0.071

good luck! : d