Question

3 sinx + 2 cosx = 3.

Answer 1

$3\sin x+2\cos x=3$

Поделим все части на корень из суммы квадратов коэффициентов перед тригонометрическими функциями.

$\sqrt{3^2+2^2}=\sqrt{9+4}=\sqrt{13}$

$\dfrac{3}{\sqrt{13}}\sin x+\dfrac{2}{\sqrt{13}}\cos x=\dfrac{3}{\sqrt{13}}$

Сделали это для того, чтобы теперь наш корень из суммы квадратов коэффициентов был равен единице. Проверим:

$\sqrt{\left(\dfrac{3}{\sqrt{13}}\right)^2+\left(\dfrac{2}{\sqrt{13}}\right)^2}=\sqrt{\dfrac{9+4}{13}}=\sqrt{\dfrac{13}{13}}=\sqrt{1}=1$

Так как это верное равенство, значит, числа $\dfrac{3}{\sqrt{13}}$ и $\dfrac{2}{\sqrt{13}}$ лежат на единичной окружности. Соответственно, существует такой угол $\varphi$, что, например, $\sin\varphi=\dfrac{3}{\sqrt{13}}$ и $\cos\varphi=\dfrac{2}{\sqrt{13}}$. Отсюда возьмём $\varphi=\arcsin\dfrac{3}{\sqrt{13}}$.

$\sin\varphi\sin x+\cos\varphi\cos x=\dfrac{3}{\sqrt{13}}\medskip\\\cos\left(x-\varphi\right)=\dfrac{3}{\sqrt{13}}\medskip\\x-\varphi=\pm\arccos\dfrac{3}{\sqrt{13}}+2\pi n,\,n\in\mathbb{Z}\medskip\\x=\varphi\pm\arccos\dfrac{3}{\sqrt{13}}+2\pi n,\,n\in\mathbb{Z}\medskip\\x=\arcsin\dfrac{3}{\sqrt{13}}\pm\arccos\dfrac{3}{\sqrt{13}}+2\pi n,\,n\in\mathbb{Z}$

Можно наш ответ "разорвать" и привести к более благородному виду:

$\left[\begin{gathered}x=\arcsin\dfrac{3}{\sqrt{13}}+\arccos\dfrac{3}{\sqrt{13}}+2\pi n,\,n\in\mathbb{Z}\\x=\arcsin\dfrac{3}{\sqrt{13}}-\arccos\dfrac{3}{\sqrt{13}}+2\pi n,\,n\in\mathbb{Z}\end{gathered}$

$\left[\begin{gathered}x=\dfrac{\pi}{2}+2\pi n,\,n\in\mathbb{Z}\\x=2\arcsin\dfrac{3}{\sqrt{13}}-\dfrac{\pi}{2}+2\pi n,\,n\in\mathbb{Z}\end{gathered}$

Ответ. $x=\dfrac{\pi}{2}+2\pi n,\,n\in\mathbb{Z}\,;~x=2\arcsin\dfrac{3}{\sqrt{13}}-\dfrac{\pi}{2}+2\pi n,\,n\in\mathbb{Z}$

Answer 2

To solve the equation 3 sin(x) + 2 cos(x) = 3, we'll use trigonometric identities and algebraic manipulation. Let's go step by step:

Step 1: Square both sides of the equation to eliminate the square root if there is any: (3 sin(x) + 2 cos(x))^2 = 3^2 (3 sin(x))^2 + 2(3 sin(x))(2 cos(x)) + (2 cos(x))^2 = 9 9 sin^2(x) + 12 sin(x) cos(x) + 4 cos^2(x) = 9

Step 2: Use the Pythagorean identity sin^2(x) + cos^2(x) = 1: 9(1 - cos^2(x)) + 12 sin(x) cos(x) + 4 cos^2(x) = 9

Step 3: Rearrange the terms and combine like terms: 9 - 9 cos^2(x) + 4 cos^2(x) + 12 sin(x) cos(x) = 9 -5 cos^2(x) + 12 sin(x) cos(x) = 0

Step 4: Divide the whole equation by -5: cos^2(x) - (12/5) sin(x) cos(x) = 0

Step 5: Use the double angle identity 2sin(x)cos(x) = sin(2x): cos^2(x) - (6/5) sin(2x) = 0

Step 6: Use the identity cos^2(x) = 1 - sin^2(x): (1 - sin^2(x)) - (6/5) sin(2x) = 0

Step 7: Rearrange the terms and bring everything to one side: sin^2(x) + (6/5) sin(2x) - 1 = 0

Now, we have a quadratic-like equation in terms of sin(x) and sin(2x). Let's make a substitution to simplify further. Let's assume y = sin(x), then sin(2x) = 2sin(x)cos(x) = 2y * √(1 - y^2).

The equation becomes: y^2 + (6/5)(2y * √(1 - y^2)) - 1 = 0

Simplify further: y^2 + (12/5)y√(1 - y^2) - 1 = 0

At this point, solving this equation for y (sin(x)) requires numerical methods, and it doesn't have a simple closed-form solution. You can use numerical techniques like Newton's method, bisection method, or graphing the equation to find approximate solutions for y (sin(x)).

Once you find the approximate values for sin(x), you can find the corresponding values of cos(x) using the identity cos^2(x) + sin^2(x) = 1.

Keep in mind that the solutions might not be exact values due to the nature of the equation, and they might involve decimal approximations.