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

Предыдущий
вопрос Следующий
вопрос

To solve this equation, we can use the pythagorean identity sin^2(x) + cos^2(x) = 1.

Substitute sin^2(x) = 1 - cos^2(x) into the equation:

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

Distribute the 2:

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

Combine like terms:

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

Divide all terms by -1:

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

Now, use the double angle identity for cosine: cos(2x) = 2cos^2(x) - 1

Substitute 2cos^2(x) - 1 for cos(2x) in the equation:

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

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

Rearranging the terms:

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

Now, use the double angle identity for sine: sin(2x) = 2sin(x)cos(x)

Substitute sin(2x) for 2sin(x)cos(x) in the equation:

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

Therefore, the solution to the equation 2sin^2(x) - 2sin(x)cos(x) - 2cos^2(x) = 1 is sin(2x) + 2cos(2x) = -1.