Question

1) 4cos3x+2sin3x=-52) 3 tg(x/2-п/3)=63) sin^2x+cos^2 2x+sin^2 3x=1.5помогите срочно решить

Answer 1

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Answer 2

Equation 1: 4cos(3x) + 2sin(3x) = -5

To solve this equation, we can use trigonometric identities to simplify it. Let's start by using the identity sin^2(x) + cos^2(x) = 1:

4cos(3x) + 2sin(3x) = -5 4cos(3x) + 2sin(3x) = -5 * (sin^2(x) + cos^2(x))

Expanding the right side of the equation:

4cos(3x) + 2sin(3x) = -5sin^2(x) - 5cos^2(x)

Now, let's use the identity sin^2(x) + cos^2(x) = 1:

4cos(3x) + 2sin(3x) = -5(1)

Simplifying further:

4cos(3x) + 2sin(3x) = -5

Now, we can use the double-angle formulas for sine and cosine:

cos(3x) = cos^2(x) - sin^2(x) sin(3x) = 2sin(x)cos(x)

Substituting these values into the equation:

4(cos^2(x) - sin^2(x)) + 2(2sin(x)cos(x)) = -5

Expanding and simplifying:

4cos^2(x) - 4sin^2(x) + 4sin(x)cos(x) = -5

Using the identity cos^2(x) = 1 - sin^2(x):

4(1 - sin^2(x)) - 4sin^2(x) + 4sin(x)cos(x) = -5

Simplifying further:

4 - 4sin^2(x) - 4sin^2(x) + 4sin(x)cos(x) = -5

Combining like terms:

8sin^2(x) - 4sin(x)cos(x) - 9 = 0

Now, we have a quadratic equation in terms of sin(x). Let's solve it using the quadratic formula:

sin(x) = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 8, b = -4cos(x), and c = -9.