Question

Sin15x+√3cos15x-2=0 ​

Answer 1

Пошаговое объяснение:

$2(\frac{1}{2} \times \sin15x + \frac{ \sqrt{3} }{2} \times \cos15x ) - 2 = 0$

$2( \cos\frac{\pi}{3} \sin15x + \frac{ \sqrt{3} }{2} \times \cos15x) - 2 = 0$

$2( \cos\frac{\pi}{3} \sin15x + sin\frac{\pi}{3} \cos15x) - 2 = 0$

$2 \sin(15x + \frac{\pi}{3} ) - 2 = 0$

$2 \sin(15x + \frac{\pi}{3} ) = 0 + 2$

$2 \sin(15x + \frac{\pi}{3} ) = 2$

$\sin(15x + \frac{\pi}{3} ) = 1$

$15x + \frac{\pi}{3} = \frac{\pi}{2 } + 2k\pi. \: k \: e \: z$

$15x = \frac{\pi}{2} - \frac{\pi}{3} + 2k\pi. \: k \: e \: z$

$15x = \frac{\pi}{6} + 2k\pi. \: k \: e \: z$

$x = \frac{\pi}{90} + \frac{2k\pi}{15} . \: k \: e \: z$

Answer 2

To solve the equation sin(15x) + √3cos(15x) - 2 = 0, we can use trigonometric identities and techniques. Let's break it down step by step.

First, let's rewrite the equation using the angle addition formula for cosine:

sin(15x) + √3cos(15x) - 2 = 0 sin(15x) + √3cos(15x) = 2

Next, let's square both sides of the equation to eliminate the square root:

[sin(15x) + √3cos(15x)]^2 = 2^2 sin^2(15x) + 2√3sin(15x)cos(15x) + 3cos^2(15x) = 4

Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we can simplify further:

1 - cos^2(15x) + 2√3sin(15x)cos(15x) + 3cos^2(15x) = 4 4cos^2(15x) - 2√3sin(15x)cos(15x) - 3 = 0

Now, let's substitute u = cos(15x) to simplify the equation:

4u^2 - 2√3sin(15x)u - 3 = 0

This is now a quadratic equation in terms of u. We can solve it using the quadratic formula:

u = (-b ± √(b^2 - 4ac)) / (2a)

where a = 4, b = -2√3sin(15x), and c = -3.

Solving for u will give us the values of cos(15x), and then we can find the corresponding values of 15x.

However, please note that finding the exact solutions for this equation involves complex trigonometric calculations, and it may not be possible to find a simple closed-form solution. In such cases, numerical methods or approximations may be more suitable for finding solutions.