(2sin(x+П/6)-1)(2tgx+1)=0

Предыдущий
вопрос Следующий
вопрос

To find the solutions to the equation:

(2sin(x+π/6) - 1)(2tan(x) + 1) = 0

We set each factor equal to zero and solve for x:

1) 2sin(x+π/6) - 1 = 0
2sin(x+π/6) = 1
sin(x+π/6) = 1/2

Using the reference angle of π/6 in the first quadrant where sine is positive, we have:
x+π/6 = π/6 + 2πn or x+π/6 = 5π/6 + 2πn

This simplifies to:
x = 0 + 2πn or x = π/2 + 2πn

2) 2tan(x) + 1 = 0
2tan(x) = -1
tan(x) = -1/2

Using the reference angle of π/6 in the fourth quadrant, we have:
x = 5π/6 + πm

So the solutions to the equation are:
x = 0 + 2πn, π/2 + 2πn, 5π/6 + πm where n and m are integers.