Question

2sin^2x-5sinx*cosx=cos^2x-2

Answer 1

Ответ:

x={π/4+kπ; arctg(1/4)+kπ}, k∈Z

Объяснение:

2sin²x-5sinxcosx=cos²x-2

2sin²x-5sinxcosx=cos²x-2(cos²x+sin²x)

2sin²x-5sinxcosx=cos²x-2cos²x-2sin²x

4sin²x-5sinxcosx+cos²x=0

(4sin²x-5sinxcosx+cos²x)/cos²x=0

4tg²x-5tgx+1=0

tgx=y

4y²-5y+1=0

4y²-4y-y+1=0

4y(y-1)-(y-1)=0

(y-1)(4y-1)=0

1) y-1=0

y=1

tgx=1

x=π/4+kπ, k∈Z

2) 4y-1=0

y=1/4

tgx=1/4

x=arctg(1/4)+kπ, k∈Z

Answer 2

To solve the equation 2sin^2(x) - 5sin(x)cos(x) = cos^2(x) - 2, you can use trigonometric identities and algebraic techniques to simplify and solve for x. Here's the step-by-step solution:

1. Start by using the trigonometric identity sin^2(x) + cos^2(x) = 1 to replace cos^2(x) in the equation:

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

2. Rearrange the equation to isolate the trigonometric terms on one side:

2sin^2(x) - 5sin(x)cos(x) - 1 = 0

3. Now, we can factor out sin(x) from the first two terms:

sin(x)(2sin(x) - 5cos(x)) - 1 = 0

4. Next, we can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the second term:

sin(x)(sin(2x) - 5cos(x)) - 1 = 0

5. Now, we have a quadratic equation in terms of sin(x):

sin(x)(sin(2x) - 5cos(x)) - 1 = 0

6. To solve this equation, set each factor equal to zero:

a) sin(x) = 0 b) sin(2x) - 5cos(x) = 0

Let's solve each equation separately:

a) sin(x) = 0 x = 0, π, 2π, ...

b) sin(2x) - 5cos(x) = 0

Use the identity sin(2x) = 2sin(x)cos(x) to rewrite the equation:

2sin(x)cos(x) - 5cos(x) = 0

Now, factor out cos(x):

cos(x)(2sin(x) - 5) = 0

Set each factor equal to zero:

i) cos(x) = 0 x = π/2, 3π/2, 5π/2, ...

ii) 2sin(x) - 5 = 0 2sin(x) = 5 sin(x) = 5/2

However, there is no real solution to sin(x) = 5/2 because the sine function is bounded between -1 and 1. So, there are no solutions for this part.

Therefore, the solutions to the original equation 2sin^2(x) - 5sin(x)cos(x) = cos^2(x) - 2 are: x = 0, π, 2π, ... (from sin(x) = 0) x = π/2, 3π/2, 5π/2, ... (from cos(x) = 0)

These are the values of x that satisfy the given equation.