Question

7sin2x=sin7x-sin3x. ​

Answer 1

$7sin2x=sin7x-sin3x$

правую сторону преобразовываем в произведение:

$7sin2x=2cos5x*sin2x$

всё переносим в левую сторону и выводим за скобки sin2x

$7sin2x-2cos5x*sin2x=0\\sin2x(7-2cos5x)=0$

приравниваем к нулю оба множителя:

$7-2cos5x=0\\2cos5x=7\\cos5x=\frac{7}{2}= 3.5\\x \in \emptyset$

не имеет решений, потому что область значений cosx = [-1;1]

$sin2x=0\\2x=\pi k, k \in Z \ (:2) \\x=\frac{\pi k}{2} , k \in Z$

Answer 2

To solve the trigonometric equation $7\sin(2x) = \sin(7x) - \sin(3x)$, we'll use trigonometric identities and algebraic manipulation to simplify it and find the values of $x$ that satisfy the equation.

Step 1: Apply trigonometric identities We'll use the following trigonometric identities:

1. $\sin(A + B) = \sin A \cos B + \cos A \sin B$
2. $\sin(A - B) = \sin A \cos B - \cos A \sin B$

Step 2: Rewrite $\sin(7x)$ and $\sin(3x)$ using the above identities $\sin(7x) = \sin(5x + 2x) = \sin(5x)\cos(2x) + \cos(5x)\sin(2x)$ $\sin(3x) = \sin(5x - 2x) = \sin(5x)\cos(2x) - \cos(5x)\sin(2x)$

Step 3: Substitute the rewritten expressions into the original equation $7\sin(2x) = \left(\sin(5x)\cos(2x) + \cos(5x)\sin(2x)\right) - \left(\sin(5x)\cos(2x) - \cos(5x)\sin(2x)\right)$

Step 4: Cancel out common terms $7\sin(2x) = 2\cos(5x)\sin(2x)$

Step 5: Solve for $\sin(2x)$ Divide both sides by $2\cos(5x)$: $\sin(2x) = \frac{7\sin(2x)}{2\cos(5x)}$

Step 6: Consider possible values We must be cautious when dividing by $\cos(5x)$ since $\cos(5x)$ can be equal to zero, which would make the equation undefined. To avoid this, we need to ensure that $\cos(5x) \neq 0$.

Step 7: Solve for $x$ Let's solve for $x$ by considering two cases:

Case 1: $\sin(2x) = 0$ If $\sin(2x) = 0$, then $2x = n\pi$ where $n$ is an integer. Thus, $x = \frac{n\pi}{2}$.

Case 2: $\cos(5x) \neq 0$ If $\cos(5x) \neq 0$, then we can divide by $\cos(5x)$: $\sin(2x) = \frac{7\sin(2x)}{2\cos(5x)}.$

Now, we can subtract $\frac{7\sin(2x)}{2\cos(5x)}$ from both sides: $0 = \sin(2x) - \frac{7\sin(2x)}{2\cos(5x)}.$

To simplify further, find a common denominator for the terms on the right-hand side: $0 = \frac{2\cos(5x)\sin(2x) - 7\sin(2x)}{2\cos(5x)}.$

Now, factor out $\sin(2x)$ from the numerator: $0 = \frac{\sin(2x)(2\cos(5x) - 7)}{2\cos(5x)}.$

Now, we have two possibilities:

1. $\sin(2x) = 0$, which corresponds to the solutions from Case 1.
2. $2\cos(5x) - 7 = 0$

Solving for $x$ in the second case: $2\cos(5x) = 7$ $\cos(5x) = \frac{7}{2}$

However, this equation has no real solutions. The range of the cosine function is $[-1, 1]$, and $\frac{7}{2}$ is outside this range, indicating there are no real values of $x$ that satisfy this equation.

So, the solutions to the original equation $7\sin(2x) = \sin(7x) - \sin(3x)$ are $x = \frac{n\pi}{2}$, where $n$ 0 0