Question

2 sin((п/4)-x)=cos(3x-(п/4))

Answer 1

Пусть $u=\frac{\pi}{4}-x$;

$2\sin u=\cos(\frac{3\pi}{4}-3u-\frac{\pi}{4})=\sin 3u$;

Рассмотрим два случая:

1) $\sin u \neq 0$

$2=\frac{\sin 3u}{\sin u}\Leftrightarrow 1= \frac{\sin 3u-\sin u}{\sin u}$

Воспользуемся формулой $\boxed{\sin x-\sin y=2\sin \frac{x-y}{2}\cos \frac{x+y}{2}}$;

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Здесь ее краткий вывод:

Пусть $x=\alpha+\beta$, а $y=\alpha-\beta$; Такие числа легко найти: $\alpha=\frac{x+y}{2},\; \beta=\frac{x-y}{2}$;

Получаем $\sin(\alpha+\beta)-\sin(\alpha-\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha-\sin\alpha\cos\beta+\sin\beta\cos\alpha$

В итоге $\sin(\alpha+\beta)-\sin(\alpha-\beta)=2\sin\beta\cos\alpha=2\sin\frac{x-y}{2}\cos\frac{x+y}{2}$;

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Итак, $1=\frac{2\sin u\cos2u}{\sin u}=2\cos 2u \Leftrightarrow \cos 2u=\frac{1}{2}\Leftrightarrow u=\pm \frac{\pi}{6}+\pi k,\k\in\mathbb{Z}$

Откуда $x=\frac{\pi}{12}+\pi k,\; x= \frac{5\pi}{12}+\pi k$

2) $\sin u =0$

Тогда $u=\pi t$  и $\sin 3u=\sin 3\pi t =0$. То есть получили равенство.

Ответ: $x=\frac{\pi}{12}+\pi k,\; x= \frac{5\pi}{12}+\pi k, x=\frac{\pi}{4}-\pi k$

Answer 2

To solve the equation 2sin(π/4 - x) = cos(3x - π/4), we can use trigonometric identities to simplify and manipulate the equation. Let's go step by step:

1. Start by expanding the trigonometric functions using the angle subtraction formula:

   2sin(π/4 - x) = cos(3x - π/4)

   2sin(π/4)cos(x) - 2cos(π/4)sin(x) = cos(3x)cos(π/4) + sin(3x)sin(π/4)

   √2(cos(x) - sin(x)) = (√2/2)cos(3x) + (√2/2)sin(3x)

2. Divide both sides of the equation by √2:

   cos(x) - sin(x) = (1/2)cos(3x) + (1/2)sin(3x)

3. Rearrange the terms on the right-hand side:

   cos(x) - sin(x) - (1/2)cos(3x) - (1/2)sin(3x) = 0

4. Combine like terms:

   cos(x) - (1/2)cos(3x) - sin(x) - (1/2)sin(3x) = 0

5. Apply the cosine sum and difference identity (cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)):

   cos(x) - (1/2)(cos(x)cos(2x) - sin(x)sin(2x)) - sin(x) - (1/2)(sin(x)cos(2x) + cos(x)sin(2x)) = 0

6. Distribute and combine like terms:

   cos(x) - (1/2)cos(x)cos(2x) + (1/2)sin(x)sin(2x) - sin(x) - (1/2)sin(x)cos(2x) - (1/2)cos(x)sin(2x) = 0

7. Combine the terms with the same trigonometric functions:

   cos(x) - (1/2)cos(x)cos(2x) - (1/2)sin(x)cos(2x) - sin(x) + (1/2)sin(x)sin(2x) - (1/2)cos(x)sin(2x) = 0

8. Factor out cos(2x) and sin(2x):

   cos(x) - sin(x) - (1/2)(cos(x) + sin(x))(cos(2x) - sin(2x)) = 0

9. Apply the Pythagorean identity (sin^2(A) + cos^2(A) = 1):

   cos(x) - sin(x) - (1/2)(cos(x) + sin(x))(1 - 2sin^2(x)) = 0

10. Expand the expression (cos(x) + sin(x))(1 - 2sin^2(x)):

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

11. Combine like terms:

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

12. Simplify further:

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

    -8sin^3(x) + 4sin^2(x) = 0

13. Factor out sin^2(x):

    -4sin^2(x)(2sin(x) - 1) = 0

14. Set each factor equal to zero:

    -4sin^2(x) = 0 or 2sin(x) - 1 = 0

15. Solve the first equation:

    -4sin^2(x) = 0

    sin^2(x) = 0

    sin(x) = 0

    x = 0, π

16. Solve the second equation:

    2sin(x) - 1 = 0

    2sin(x) = 1

    sin(x) = 1/2

    x = π/6, 5π/6

Therefore, the solutions to the equation 2sin(π/4 - x) = cos(3x - π/4) are x = 0, π, π/6, and 5π/6.