Question

4 sin 2x+cos^2 2x = 4

Answer 1

$4\sin 2x + \cos^2 2x = 4\\\\4\sin 2x + 1 -\sin^2 2x - 4 = 0\\\\-\sin^22x + 4\sin 2x - 3 = 0\ \ \ \ \ \Big| \cdot (-1)\\\\\sin^22x - 4\sin 2x + 3 = 0$

Введём замену:  $t = \sin 2x\ ,\ t\in[-1; 1]$ .

$t^2 - 4t + 3 = 0$

По теореме Виета:

$\begin{equation*}\begin{cases}t_{1}t_{2} = 3\\t_{1} + t_{2} = 4\end{cases}\end{equation*}\ \ \ \ \ \Big| t = 3; t = 1\ \ \ \Rightarrow\ \boxed{\bf{t = 1}}$

Корень 3 не находится в области допустимых значений переменной $t$, поэтому он не является решением.

Обратная замена:

$\sin 2x = 1\\\\2x = \dfrac{\pi}{2} + 2\pi k\\\\\\\boxed{\bf{x = \dfrac{\pi}{4} + \pi k}}}\ ,\ k\in\mathbb{Z}$

Answer 2

To solve the equation 4sin(2x) + cos^2(2x) = 4, we can manipulate the equation to express it in terms of a single trigonometric function.

Let's start by using the trigonometric identity cos^2(x) = 1 - sin^2(x):

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

Rearranging the terms:

sin^2(2x) + 4sin(2x) - 3 = 0

Now, let's make a substitution: let u = sin(2x):

u^2 + 4u - 3 = 0

We can now solve this quadratic equation for u:

(u + 3)(u - 1) = 0

This gives us two possible values for u:

u + 3 = 0 or u - 1 = 0

u = -3 or u = 1

Now, substitute back u = sin(2x):

sin(2x) = -3 or sin(2x) = 1

However, the sine function can only take values between -1 and 1. Therefore, sin(2x) = -3 does not have a valid solution.

For sin(2x) = 1, we can find the solutions by solving for 2x:

2x = arcsin(1) = π/2 + 2πn (n is an integer)

Dividing by 2:

x = π/4 + πn (n is an integer)

So the solutions to the equation 4sin(2x) + cos^2(2x) = 4 are:

x = π/4 + πn (n is an integer)