Question

(√3)*tg+3=0 [п/3;3п/2]

Answer 1

$\sqrt{3} tgx+3=0 \\ tgx=- \frac{3}{\sqrt{3}} \\ tgx=-\sqrt{3} \\ x=- \frac{\pi}{3} + \pi n,n \in Z \\ \\ n=1;\,\,\,\,x= \frac{2 \pi }{3}$

Answer 2

To solve the equation (√3)*tan(x) + 3 = 0 on the interval [π/3, 3π/2], we can use the following steps:

Step 1: Rewrite the equation (√3)*tan(x) + 3 = 0 as tan(x) = -3/√3.

Step 2: Simplify the right side of the equation. -3/√3 can be rationalized by multiplying the numerator and denominator by √3, giving -3√3/3, which simplifies to -√3.

So, we have tan(x) = -√3.

Step 3: Find the angles where the tangent function is equal to -√3. One such angle is π/3.

Step 4: Determine the reference angle. The reference angle is the positive acute angle formed between the terminal side of the angle and the x-axis. In this case, the reference angle is π/3 since it is within the specified interval.

Step 5: Add the reference angle to the multiples of π to get the complete set of solutions within the given interval.

The solutions for x in the interval [π/3, 3π/2] are:

x = π/3 + kπ, where k is an integer such that π/3 ≤ x ≤ 3π/2.

In other words, the solutions are x = π/3, 4π/3, 7π/3, 10π/3, etc., as long as these values fall within the specified interval.