Question

Sin x + 2 cos x - 1 = 0​

Answer 1

$\sin x+2\cos x-1=0\\ \\ 2\sin \frac{x}{2}\cos \frac{x}{2}+2\Big(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\Big)-\sin^2\frac{x}{2}-\cos^2\frac{x}{2}=0\\ \\ 2\sin\frac{x}{2}\cos \frac{x}{2}+\cos^2\frac{x}{2}-3\sin^2\frac{x}{2}=0~~~~~~~~~\Bigg|:\Big(-\cos^2\frac{x}{2}\ne 0\Big)$

$3{\rm tg}^2\frac{x}{2}-2{\rm tg}\,\frac{x}{2}-1=0$

Решим последнее уравнение методом разложения на множители. Для этого левую часть уравнения можно представить в виде:

$3{\rm tg}^2\,\frac{x}{2}-3{\rm tg}\,\frac{x}{2}+{\rm tg}\,\frac{x}{2}-1=0\\ \\ 3{\rm tg}\,\frac{x}{2}\Big({\rm tg}\,\frac{x}{2}-1\Big)+{\rm tg}\,\frac{x}{2}-1=0\\ \\ \Big({\rm tg}\,\frac{x}{2}-1\Big)\Big(3{\rm tg}\,\frac{x}{2}+1\Big)=0\\ \\ {\rm tg}\,\frac{x}{2}-1=0\\ \\ \dfrac{x}{2}=\dfrac{\pi}{4}+\pi n,n \in \mathbb{Z}\\ \\ \boxed{x_1=\frac{\pi}{2}+2\pi n,n \in \mathbb{Z}}$

$3{\rm tg}\, \frac{x}{2}+1=0\\ \\ \dfrac{x}{2}=-{\rm arctg}\, \frac{1}{3}+\pi n,n \in \mathbb{Z}\\ \\ \boxed{x_2=-2{\rm arctg}\, \frac{1}{3}+2\pi n,n \in \mathbb{Z}}$

Answer 2

To solve the equation sin(x) + 2cos(x) - 1 = 0, you can use trigonometric identities to simplify and solve for x. Here's how you can do it:

1. Start by expressing cos(x) in terms of sin(x) using the Pythagorean identity: cos(x) = √(1 - sin^2(x))

2. Substitute this expression for cos(x) into the equation: sin(x) + 2√(1 - sin^2(x)) - 1 = 0

3. Now, let's isolate sin(x) on one side of the equation: sin(x) + 2√(1 - sin^2(x)) = 1

4. Square both sides of the equation to eliminate the square root: (sin(x) + 2√(1 - sin^2(x)))^2 = 1^2

5. Expand the left side of the equation: sin^2(x) + 4sin(x)√(1 - sin^2(x)) + 4(1 - sin^2(x)) = 1

6. Move all terms to one side of the equation to set it to zero: sin^2(x) + 4sin(x)√(1 - sin^2(x)) + 4(1 - sin^2(x)) - 1 = 0

7. Now, let's make a substitution to simplify the equation. Let u = sin(x): u^2 + 4u√(1 - u^2) + 4(1 - u^2) - 1 = 0

8. This is a quadratic equation in u. Solve it for u.

   Let's rewrite the equation in terms of the square root: u^2 + 4u√(1 - u^2) + 4(1 - u^2) - 1 = 0

   Move the terms involving the square root to the right side: u^2 + 4(1 - u^2) = 1 - 4u√(1 - u^2)

   Rearrange the terms: u^2 + 4 - 4u^2 = 1 - 4u√(1 - u^2)

   Combine like terms: 4 - 3u^2 = 1 - 4u√(1 - u^2)

   Move all terms to one side: 4u√(1 - u^2) - 3u^2 - 3 = 0

9. Now, you can solve this equation for u. You may need to use numerical methods or a calculator to find the solution for u.

Once you have found the values of u, you can use them to find the corresponding values of x by taking the arcsin (inverse sine) of u. Remember that the solution to the original equation may involve multiple values of x, as trigonometric equations often have periodic solutions.