Question

2tgxcosx+1=2cosx+tgx

Answer 1

Решение задания приложено

Answer 2

To solve the equation 2tg(x)cos(x) + 1 = 2cos(x) + tg(x), we can follow these steps:

Step 1: Bring all the terms to one side of the equation: 2tg(x)cos(x) + 1 - (2cos(x) + tg(x)) = 0

Step 2: Simplify the left side of the equation: 2tg(x)cos(x) - 2cos(x) - tg(x) + 1 = 0

Step 3: Factor out common terms: tg(x)(2cos(x) - 1) - (2cos(x) - 1) = 0

Step 4: Factor out (2cos(x) - 1): (2cos(x) - 1)(tg(x) - 1) = 0

Now, we have two possible cases:

Case 1: 2cos(x) - 1 = 0 Solve for cos(x): 2cos(x) = 1 cos(x) = 1/2

The solutions for this case are the angles where the cosine function equals 1/2: x = π/3 + 2kπ (where k is an integer), or x = 5π/3 + 2kπ (where k is an integer).

Case 2: tg(x) - 1 = 0 Solve for tg(x): tg(x) = 1

The solutions for this case are the angles where the tangent function equals 1: x = π/4 + kπ (where k is an integer).

So, the complete solution to the equation is: x = π/3 + 2kπ, x = 5π/3 + 2kπ, or x = π/4 + kπ, where k is an integer.