Question

1.)2sinx=√2 2.)tgx-1=0

Answer 1

Во вложении.............

Answer 2

Let's solve each equation one by one:

1. 2sin(x) = √2

To solve for x, we need to isolate sin(x) first. Divide both sides of the equation by 2:

sin(x) = √2 / 2

Now, √2 / 2 is a common value in trigonometry. It represents the sine of 45 degrees or π/4 radians. Therefore:

sin(x) = sin(π/4)

However, sin(x) can also be equal to √2 / 2 in the second and third quadrants. These angles are π - π/4 = 3π/4 and 2π + π/4 = 9π/4.

So, the solutions for x are: x = π/4 + 2kπ, where k is an integer (for angles in the first quadrant) x = 3π/4 + 2kπ, where k is an integer (for angles in the second quadrant) x = 9π/4 + 2kπ, where k is an integer (for angles in the third quadrant)

2. tan(x) - 1 = 0

To solve for x, we need to isolate the tangent function. Add 1 to both sides of the equation:

tan(x) = 1

Again, this is a common value in trigonometry. It represents the tangent of 45 degrees or π/4 radians. So:

x = π/4 + kπ, where k is an integer

These are the solutions for the equation. Each solution represents an angle at which the tangent of that angle is equal to 1. Keep in mind that the tangent function repeats itself every π radians, so adding kπ accounts for all possible angles that satisfy the equation.