Question

1) sin(x+пи/4)=-1 2) tg3x= -корень из 3/ 3

Answer 1

1) $x+\pi/4=-\pi/2+2\pi k,\ k\in\mathbf{Z}$

$x=-3\pi/4+2\pi k,\ k\in\mathbf{Z}$

 

2) $3x=-\pi/6+\pi k,\ k\in\mathbf{Z}$

$x=-\pi/18+\pi k/3,\ k\in\mathbf{Z}$

 

Answer 2

1. To solve the equation sin(x + π/4) = -1, we can first determine the angle whose sine is -1. Since the sine function is negative in the third and fourth quadrants, the angle we are looking for is π - π/2, or 3π/2.

Therefore, we can write x + π/4 = 3π/2 + 2kπ or x + π/4 = 5π/2 + 2kπ, where k is any integer. Solving for x in each equation, we get:

x = 5π/2 - π/4 + 2kπ = 11π/4 + 2kπ or x = 3π/2 - π/4 + 2kπ = 5π/4 + 2kπ

So the solutions are x = 11π/4 + 2kπ or x = 5π/4 + 2kπ, where k is any integer.

2. To solve the equation tan(3x) = -√3/3, we can use the tangent half-angle formula:

tan(2θ) = 2tan(θ) / (1 - tan^2(θ))

Setting θ = 3x/2, we get:

tan(3x) = 2tan(3x/2) / (1 - tan^2(3x/2))

Substituting y = tan(3x/2), we have:

tan(3x) = 2y / (1 - y^2)

Solving for y, we get:

y = -√3/3 or y = √3

Since y = tan(3x/2), we can find 3x/2 as follows:

3x/2 = arctan(-√3/3) + kπ or 3x/2 = arctan(√3) + kπ, where k is any integer.

Using a calculator or reference table, we can find that arctan(-√3/3) = -π/6 + 2πk or arctan(√3) = π/3 + 2πk, where k is any integer.

Substituting these values, we get:

3x/2 = -π/6 + 2πk or 3x/2 = π/3 + 2πk

Solving for x, we get:

x = -π/9 + 4πk/3 or x = π/2 + 2πk/3, where k is any integer.