Question

Lg^2 (tg^2x)+lg(cosx)=lg(sinx)

Answer 1

lg^2 (tg^2 x) + lg (cos x) = lg (sin x)
Область определения:
{ sin x > 0
{ cos x > 0
x € (2pi*k; pi/2+2pi*k)
[lg (tg^2 x)]^2 = lg (sin x) - lg (cos x)
[2lg (tg x)]^2 = lg (sin x/cos x) = lg (tg x)
4[lg (tg x)]^2 - lg (tg x) = 0
lg (tg x)*(4lg (tg x) - 1) = 0
1) lg (tg x) = 0
tg x = 1
x1 = pi/4 + pi*n
С учетом Обл. Опр. x1 = pi/4 + 2pi*n
2) 4lg (tg x) - 1 = 0
lg(tg x) = 1/4 = lg(10^(1/4))
tg x = 10^(1/4) = корень 4 степени из 10
x2 = arctg (10^(1/4)) + pi*k
С учетом Обл. Опр. x2 = arctg(10^(1/4)) + 2pi*k

Answer 2

To solve the equation lg^2(tg^2x) + lg(cosx) = lg(sinx), let's break it down step by step:

1. Start by converting the equation from logarithmic form to exponential form. Recall that lg(x) is equivalent to log base 10 of x. The exponential form is 10^y = x, where y is the logarithmic expression and x is the value.

   Therefore, we can rewrite the equation as follows: 10^(lg^2(tg^2x)) * 10^(lg(cosx)) = 10^(lg(sinx))

2. Apply the properties of logarithms to simplify the equation. Recall that lg^2(tg^2x) = (lg(tg^2x))^2 and 10^(lg(x)) = x. We can simplify the equation as follows: (tg^2x)^2 * cosx = sinx

3. Rewrite tg^2x as (sinx/cosx)^2: ((sinx/cosx)^2)^2 * cosx = sinx

4. Simplify the equation further: (sin^2x/cos^2x) * cosx = sinx

5. Cancel out the common factors of cosx: sin^2x = sinx

6. Rearrange the equation and factor out sinx: sinx * (sinx - 1) = 0

7. Now, set each factor equal to zero and solve for x:

   • sinx = 0
   • sinx - 1 = 0

   Solving these equations, we find two sets of solutions:

   • For sinx = 0, the solutions are x = 0 + nπ (where n is an integer).
   • For sinx - 1 = 0, the solution is x = π/2.

Therefore, the solutions to the equation lg^2(tg^2x) + lg(cosx) = lg(sinx) are x = 0 + nπ and x = π/2, where n is an integer.