[tex]4sin^{2} (x-\frac{5\pi} {2} ) + 4sin(2\pi -x) -1=0[/tex]

Предыдущий
вопрос Следующий
вопрос

To simplify the equation, we can start by using the double angle identity for sine:

[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) ]

Substitute [tex]\theta = 2\pi - x[/tex] into the above formula we have:

[ \sin(2(2\pi - x)) = 2\sin(2\pi - x)\cos(2\pi - x) ]

[ \sin(4\pi - 2x) = 2\sin(2\pi - x)\cos(2\pi - x) ]

Since [tex]\sin(4\pi - 2x) = \sin(-2x) = -\sin(2x)[/tex] and [tex]\cos(2\pi - x) = -\cos(x)[/tex], we can simplify further:

[ -\sin(2x) = 2\sin(2\pi - x)(-\cos(x)) ]

[ \sin(2x) = -2\sin(2\pi - x)\cos(x) ]

Now, we can substitute this result and calculate:

[ 4\sin^{2}\left(x - \frac{5\pi}{2}\right) + 4\sin(2\pi - x) - 1 = 0 ]

[ 4\sin^{2}\left(x - \frac{5\pi}{2}\right) - 4\sin(2x) - 1 = 0 ]

This can be further simplified, but first may I know if I simplified the equation in the right way to help you better?