Question

Решите уравнение(sin2x+cosx)(кореньиз 3+корень из 3tgx)=0 помогите решить плиз!(

Answer 1

Sin2x+cosx=0
2sinxcosx+cosx=0
cosx(2sinx+1)=0
cosx=0⇒x=π/2+πn,n∈z
sinx=-1/2⇒x=(-1)^(k+1)*π/6+πk,k∈z
√3+√3tgx=0
√3tgx=-√3
tgx=-1⇒x=-π/4+πm,m∈z

Answer 2

Solving the Equation

To solve the equation (sin2x + cosx)(√3 + √3tgx) = 0, we need to find the values of x that satisfy this equation.

Let's break down the equation and solve it step by step.

1. First, let's consider the first factor: sin2x + cosx. This is a trigonometric expression that involves sine and cosine functions. We can simplify it using trigonometric identities.

Using the double-angle identity for sine, we have: sin2x = 2sinxcosx.

Substituting this back into the equation, we get: 2sinxcosx + cosx.

Factoring out cosx, we have: cosx(2sinx + 1).

2. Now, let's consider the second factor: √3 + √3tgx. This involves the tangent function.

We can rewrite tgx as sinx/cosx.

Substituting this back into the equation, we get: √3 + √3sinx/cosx.

Simplifying further, we have: (√3cosx + √3sinx)/cosx.

3. Combining the two factors, we have: cosx(2sinx + 1)(√3cosx + √3sinx)/cosx.

Simplifying, we get: (2sinx + 1)(√3cosx + √3sinx).

4. Now, we set the equation equal to zero and solve for x:

(2sinx + 1)(√3cosx + √3sinx) = 0.

This equation will be satisfied if either of the factors is equal to zero.

Setting the first factor equal to zero, we have: 2sinx + 1 = 0.

Solving for x, we get#### Solving the Equation

To solve the equation (sin2x + cosx)(√3 + √3tgx) = 0, we need to find the values of x that satisfy this equation.

Let's break down the equation and solve it step by step.

First, let's consider the product of the two factors: (sin2x + cosx)(√3 + √3tgx).

The equation will be satisfied if either of the two factors is equal to zero. So we have two cases to consider:

Case 1: sin2x + cosx = 0 To solve this equation, we can use the trigonometric identity: sin2x = 2sinxcosx.

Substituting this identity into the equation, we get: 2sinxcosx + cosx = 0

Factoring out cosx, we have: cosx(2sinx + 1) = 0

This equation will be satisfied if either cosx = 0 or 2sinx + 1 = 0.

For cosx = 0, the solutions are x = π/2 + nπ, where n is an integer.

For 2sinx + 1 = 0, we can solve for sinx: 2sinx = -1 sinx = -1/2

The solutions for sinx = -1/2 are x = 7π/6 + 2nπ and x = 11π/6 + 2nπ, where n is an integer.

Case 2: √3 + √3tgx = 0 To solve this equation, we need to isolate tgx.

Subtracting √3 from both sides, we have: √3tgx = -√3

Dividing both sides by √3, we get: tgx = -1

The solutions for tgx = -1 are x = 3π/4 + nπ, where n is an integer.

Summary of Solutions

Combining the solutions from both cases, we have the following values of x that satisfy the equation:

- For cosx = 0: x = π/2 + nπ, where n is an integer. - For sinx = -1/2: x = 7π/6 + 2nπ and x = 11π/6 + 2nπ, where n is an integer. - For tgx = -1: x = 3π/4 + nπ, where n is an integer.

These are the solutions to the equation (sin2x + cosx)(√3 + √3tgx) = 0.

Please note that these solutions are based on the given equation and the trigonometric identities used.