Question

1/2sin4x=1 1-cos^2x=02tgx-2ctgx=3cosx=cos4cos2x=1-3cosx; 1<x<50(sinx+1)(ctg2x-√3)=0

Answer 1

Вот, все что смогла, надеюсь поможет)) 

Answer 2

Let's break down the given equations step by step:

1. 1/2sin(4x) = 1: To solve this equation, we can start by multiplying both sides by 2 to eliminate the fraction: sin(4x) = 2.

2. 1 - cos^2(x) = 0: This equation can be rewritten using the identity cos^2(x) + sin^2(x) = 1: sin^2(x) = 1 - cos^2(x).

3. 2tg(x) - 2ctg(x) = 3cos(x): To simplify this equation, let's express tangent (tg) and cotangent (ctg) in terms of sine and cosine: 2(sin(x)/cos(x)) - 2(cos(x)/sin(x)) = 3cos(x). Simplifying further: 2sin(x)^2 - 2cos(x)^2 = 3cos(x)sin(x). Using the identity sin(2x) = 2sin(x)cos(x): sin(2x) - cos(2x) = (3/2)cos(x)sin(x).

4. cos(4x)cos(2x) = 1 - 3cos(x): This equation involves multiple trigonometric functions. We can simplify it using various trigonometric identities. Let's break it down: cos(4x)cos(2x) = 1 - 3cos(x). Using the double-angle formula for cosine, cos(2x) = 2cos^2(x) - 1: cos(4x)(2cos^2(x) - 1) = 1 - 3cos(x). Expanding further: 2cos^2(x)cos(4x) - cos(4x) = 1 - 3cos(x).

5. (sin(x) + 1)(ctg(2x) - √3) = 0: This equation can be solved by considering each factor separately: a) sin(x) + 1 = 0: sin(x) = -1.

b) ctg(2x) - √3 = 0: ctg(2x) = √3. Recall that cotangent (ctg) is the reciprocal of tangent (tg), so we can rewrite it as: tg(2x) = 1/√3.

Please note that these equations involve multiple trigonometric functions and may not have simple exact solutions. To find the specific values of x that satisfy these equations within the given range of 1 < x < 50, numerical methods or a calculator may be required.