Question

2cos2x+2sinxcosx =0 ​

Answer 1

думаю вот так будет решение

Answer 2

To solve the equation 2cos(2x) + 2sin(x)cos(x) = 0, you can use trigonometric identities and algebraic manipulation. Here's how you can do it:

First, use the double-angle identity for cosine: cos(2x) = 2cos^2(x) - 1

Now, substitute this expression into the equation: 2(2cos^2(x) - 1) + 2sin(x)cos(x) = 0

Next, distribute the 2 on the left side of the equation: 4cos^2(x) - 2 + 2sin(x)cos(x) = 0

Now, add 2 to both sides of the equation to isolate the trigonometric terms: 4cos^2(x) + 2sin(x)cos(x) - 2 = 0

Now, we have a quadratic equation in terms of cos(x): 4cos^2(x) + 2sin(x)cos(x) - 2 = 0

To make it easier to solve, let's make a substitution: Let y = cos(x)

Now, rewrite the equation in terms of y: 4y^2 + 2sin(x)y - 2 = 0

Now, this is a quadratic equation in terms of y. You can solve it using the quadratic formula: y = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 4, b = 2sin(x), and c = -2. Plug in these values into the quadratic formula: y = (-2sin(x) ± √((2sin(x))² - 4(4)(-2))) / (2(4))

Now, simplify the expression inside the square root: y = (-2sin(x) ± √(4sin^2(x) + 32)) / 8

Simplify further: y = (-sin(x) ± √(sin^2(x) + 8)) / 4

Now, you have two possible solutions for y:

1. y = (-sin(x) + √(sin^2(x) + 8)) / 4
2. y = (-sin(x) - √(sin^2(x) + 8)) / 4

Remember that y = cos(x), so these are the solutions for cos(x).

So, the solutions for the original equation 2cos(2x) + 2sin(x)cos(x) = 0 are:

1. cos(x) = (-sin(x) + √(sin^2(x) + 8)) / 4
2. cos(x) = (-sin(x) - √(sin^2(x) + 8)) / 4

These are the exact solutions for the equation in terms of cos(x) and sin(x).