Question

1)3tgx=√32)2cosx+√3=0

Answer 1

3tgx=корень из 3;
tgx=корень из 3/3;
х= Пи/6+Пиn, n принадлежит целым.

2cosx+корень из 3=0;
2cosx=-корень из 3;
cosx= - корень из 3/2;
x=+-5Пи/6+2Пиk, k принадлежит целым

Answer 2

1. To solve the equation 3tgx = √3, we can use the tangent function to isolate x:

tgx = √3/3

Using the inverse tangent function (arctan) on both sides, we get:

x = arctan(√3/3)

This can be simplified using the special triangle with angles 30-60-90, where the tangent of 30 degrees is √3/3. Therefore, we have:

x = 30 degrees + k*180 degrees (where k is an integer)

So the solutions are x = 30 degrees, 210 degrees, 390 degrees, etc.

2. To solve the equation 2cosx + √3 = 0, we can isolate cosx:

2cosx = -√3

cosx = -√3/2

Using the unit circle or a reference triangle in the third quadrant, where cosx is negative and the cosine of 30 degrees is √3/2, we have:

x = 150 degrees + k*360 degrees (where k is an integer)

So the solutions are x = 150 degrees, 510 degrees, 870 degrees, etc.