Question

2cos^2x +sin2x - 4sin^2x=0

Answer 1

Ответ:

Пошаговое объяснение:

Answer 2

To solve the trigonometric equation $2\cos^2(x) + \sin(2x) - 4\sin^2(x) = 0$, we'll first try to simplify the equation and then find the values of $x$ that satisfy it.

Step 1: Use trigonometric identities to simplify the equation. Recall the double angle formulas: $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos^2(x) = 1 - \sin^2(x)$.

Now, let's substitute these identities into the equation:

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

Step 2: Combine like terms: $2 - 2\sin^2(x) + 2\sin(x)\cos(x) - 4\sin^2(x) = 0$

Step 3: Rearrange the terms to form a quadratic equation: $-6\sin^2(x) + 2\sin(x)\cos(x) + 2 = 0$

Step 4: Divide the entire equation by -2 to simplify: $3\sin^2(x) - \sin(x)\cos(x) - 1 = 0$

Step 5: Factor the equation (if possible): This equation cannot be easily factored, so we'll use the quadratic formula:

For a quadratic equation $ax^2 + bx + c = 0$, the solutions for $x$ are given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 3$, $b = -1$, and $c = -1$.

Step 6: Use the quadratic formula to find the values of $x$: $x = \frac{1 \pm \sqrt{(-1)^2 - 4(3)(-1)}}{2(3)}$

$x = \frac{1 \pm \sqrt{1 + 12}}{6}$

$x = \frac{1 \pm \sqrt{13}}{6}$

So, the solutions for $x$ are approximately: $x \approx \frac{1 + \sqrt{13}}{6}$ and $x \approx \frac{1 - \sqrt{13}}{6}$

Note: The values obtained here are approximate due to the presence of the square root of 13. If you need a numerical approximation, you can use a calculator to get decimal values.