2 sin^2 x -sin2x-2cos2x=0

Предыдущий
вопрос Следующий
вопрос

To solve the equation 2sin^2x - sin2x - 2cos2x = 0, we can use the double angle identities to rewrite sin2x and cos2x in terms of sinx and cosx.

Using the double angle identities:
sin2x = 2sinxcosx
cos2x = 2cos^2x - 1

Substitute these into the equation:
2sin^2x - 2sinxcosx - 4cos^2x + 2 = 0

Rearrange the terms:
2sin^2x - 2sinxcosx - 4cos^2x + 2 = 0
2(sin^2x - sinxcosx - 2cos^2x) + 2 = 0
2(sin^2x - 2cos^2x - sinxcosx) + 2 = 0

Factor out a negative sign:
-2(cos^2x - sinxcosx - sin^2x) - 2 = 0
-2((cosx - sinx)^2) - 2 = 0
-2(cosx - sinx)^2 - 2 = 0

Divide by -2:
(cosx - sinx)^2 + 1 = 0

Since the square of a real number is always greater than or equal to zero, the equation has no real solutions.

Therefore, the equation 2sin^2x - sin2x - 2cos2x = 0 has no real solutions.