Question

2 sin^2 x - cos 2x = 1

Answer 1

2sin²x-(cos²x-sin²x)=1
2sin²x-cos²x+sin²x=1
3sin²x-(1-sin²)=1
3sin²x-1+sin²x=1
4sin²x=2
sin²x=1/2
(sinx-1/2)(sinx+1/2)=0
sinx=+-1/√2=+-√2/2
x=π/4+πk, k∈Z

Answer 2

To solve the equation 2sin^2(x) - cos(2x) = 1, we can simplify the equation and then apply trigonometric identities to solve for x.

Let's start by using the double-angle identity for cosine: cos(2x) = 2cos^2(x) - 1

Substituting this into the equation, we have: 2sin^2(x) - (2cos^2(x) - 1) = 1

Expanding and rearranging the terms, we get: 2sin^2(x) - 2cos^2(x) + 1 = 1

Now, let's use the Pythagorean identity for sin^2(x) and cos^2(x): sin^2(x) + cos^2(x) = 1

Rearranging this equation, we have: sin^2(x) = 1 - cos^2(x)

Substituting this into the previous equation, we get: 2(1 - cos^2(x)) - 2cos^2(x) + 1 = 1

Expanding and simplifying, we have: 2 - 2cos^2(x) - 2cos^2(x) + 1 = 1

Combining like terms, we get: -4cos^2(x) + 2 = 0

Dividing by -2, we have: 2cos^2(x) - 1 = 0

Adding 1 to both sides, we get: 2cos^2(x) = 1

Dividing by 2, we have: cos^2(x) = 1/2

Taking the square root of both sides, we have: cos(x) = ±√(1/2)

Now, let's find the values of x for which cos(x) = √(1/2) and cos(x) = -√(1/2).

For cos(x) = √(1/2): x = π/4 + 2πn or x = -π/4 + 2πn

For cos(x) = -√(1/2): x = 3π/4 + 2πn or x = -3π/4 + 2πn

where n is an integer.

Therefore, the solutions to the equation 2sin^2(x) - cos(2x) = 1 are: x = π/4 + 2πn, -π/4 + 2πn, 3π/4 + 2πn, -3π/4 + 2πn, where n is an integer.