Question

14sin^2x-4cos^2x=5sin2x

Answer 1

Ответ:

вот

Объяснение:

Answer 2

To solve the trigonometric equation $14\sin^2x - 4\cos^2x = 5\sin2x$, we can use trigonometric identities to simplify and solve for $x$. Let's start by manipulating the equation step by step.

First, we'll use the double angle formula for sine: $\sin 2x = 2 \sin x \cos x$.

Substitute this into the equation:

$14\sin^2x - 4\cos^2x = 5(2\sin x \cos x)$

Now, let's work with the left-hand side of the equation:

$14\sin^2x - 4\cos^2x$

We can use the trigonometric identity $\sin^2x + \cos^2x = 1$ to substitute for $\cos^2x$:

$14\sin^2x - 4(1 - \sin^2x) = 5(2\sin x \cos x)$

Simplify the equation further:

$14\sin^2x - 4 + 4\sin^2x = 10\sin x \cos x$

Combine like terms:

$18\sin^2x - 4 = 10\sin x \cos x$

Now, we can use the identity $\cos^2x = 1 - \sin^2x$ to substitute for $\cos x \cos x$:

$18\sin^2x - 4 = 10\sin x (1 - \sin^2x)$

Distribute the $10\sin x$ term:

$18\sin^2x - 4 = 10\sin x - 10\sin^3x$

Rearrange the terms:

$10\sin^3x + 18\sin^2x - 10\sin x - 4 = 0$

Factor out a common $2\sin x$ term:

$2\sin x (5\sin^2x + 9\sin x - 2) = 0$

Now we have a quadratic factor inside the parentheses. Solve this quadratic equation for $\sin x$:

$5\sin^2x + 9\sin x - 2 = 0$

This can be factored as:

$(5\sin x - 1)(\sin x + 2) = 0$

So, $5\sin x - 1 = 0$ or $\sin x + 2 = 0$.

Solve for $\sin x$ in each case:

1. $5\sin x - 1 = 0$ $\sin x = \frac{1}{5}$

2. $\sin x + 2 = 0$ $\sin x = -2$ (This solution is not valid since the range of sine function is between -1 and 1.)

So, the valid solution is $\sin x = \frac{1}{5}$.

Finally, find the values of $x$ that satisfy $\sin x = \frac{1}{5}$. You can use inverse sine (arcsin) function:

$x = \arcsin\left(\frac{1}{5}\right) + 2n\pi$ where $n$ is an integer.

Keep in mind that trigonometric equations may have multiple solutions due to the periodic nature of trigonometric functions.