Question

2sin2x * cos2x - 1 =0​

Answer 1

$2\sin 2x \cos 2x - 1 = 0$

$\sin 4x - 1 = 0$

$\sin 4x = 1$

$4x = \dfrac{\pi}{2} + 2\pi n, \ n \in Z \ \ \ |: 4$

$x = \dfrac{\pi}{8} + \dfrac{ \pi n}{2} , \ n \in Z$

Ответ: $x = \dfrac{\pi}{8} + \dfrac{ \pi n}{2} , \ n \in Z$

Answer 2

To solve the trigonometric equation $2\sin(2x) \cos(2x) - 1 = 0$, we can use trigonometric identities and algebraic manipulation. First, let's express $\sin(2x)$ and $\cos(2x)$ in terms of basic trigonometric functions:

$\sin(2x) = 2\sin(x)\cos(x)$ $\cos(2x) = \cos^2(x) - \sin^2(x)$

Now, substitute these expressions back into the original equation:

$2\cdot (2\sin(x)\cos(x)) \cdot (\cos^2(x) - \sin^2(x)) - 1 = 0$

Simplify the equation step by step:

$4\sin(x)\cos(x)(\cos^2(x) - \sin^2(x)) - 1 = 0$

$4\sin(x)\cos(x)(\cos^2(x) - (1 - \cos^2(x))) - 1 = 0$

$4\sin(x)\cos(x)(2\cos^2(x) - 1) - 1 = 0$

$8\sin(x)\cos^3(x) - 4\sin(x)\cos(x) - 1 = 0$

Now, let's use a substitution: $u = \cos(x)$. This gives us $du = -\sin(x) dx$. Substituting these into the equation:

$8u^3 - 4u - 1 = 0$

This is a cubic equation in terms of $u$. Unfortunately, finding analytical solutions for cubic equations can be quite complex and may require numerical methods or special techniques.

If you're looking for approximate solutions, you can use numerical methods like the Newton-Raphson method or use software like calculators or mathematical software to find the roots. If you have a specific value of $x$ or a range in mind, these methods could help you find the corresponding $u$ (and thus $\cos(x)$) values.