Question

(2cosx+1) (3tgx+1)= 0

Answer 1

2cosx + 1 = 0
3tgx + 1 = 0

2cosx = -1
3tgx = -1

cosx = -0,5
tgx = -1/3

cosx = -0,5
x=arccos(-0,5) + 2pk
x = -arccos(-0,5) + 2pk
x1=2p/3 + 2pk
x2= -2p/3 + 2pk

Answer 2

To solve the equation (2cos(x) + 1)(3tan(x) + 1) = 0, we need to find the values of x that satisfy this equation. To do that, we can set each factor equal to zero and solve for x separately.

First factor: 2cos(x) + 1 = 0

Subtracting 1 from both sides: 2cos(x) = -1

Dividing by 2: cos(x) = -1/2

This means that x can take on values where the cosine of x is equal to -1/2. Looking at the unit circle, we find that the cosine is equal to -1/2 at two angles: 2π/3 and 4π/3. So, x can be:

x = 2π/3 + 2πn (where n is an integer) x = 4π/3 + 2πn (where n is an integer)

Second factor: 3tan(x) + 1 = 0

Subtracting 1 from both sides: 3tan(x) = -1

Dividing by 3: tan(x) = -1/3

This means that x can take on values where the tangent of x is equal to -1/3. The tangent function repeats every π, so we can find one angle and then add integer multiples of π to get all solutions. Using the arctan function, we find that one solution is approximately -0.32175 radians or -18.43 degrees. So, x can be:

x ≈ -0.32175 + πn (where n is an integer)

Therefore, the solutions to the equation (2cos(x) + 1)(3tan(x) + 1) = 0 are:

x = 2π/3 + 2πn (where n is an integer) x = 4π/3 + 2πn (where n is an integer) x ≈ -0.32175 + πn (where n is an integer)