Question

1) 49^x - 6-7^x-7=0 2) cos2x+sinx=0 3) 5sin^2x=3sinxcosx+4cos^2x=3 третье с решением если можно))

Answer 1

1) 49^x-6*7^x-7=0
7^2x-6*7^x-7=0
7^x=t>0
t²-6t-7=0
t1=7  t2=-1<0
7^x=7⇒x=1
2) cos2x+sinx=0
1-2sin²x+sinx=0
2sin²x-sinx-1=0 решаем как квадратное через дискриминант
D=1-4*2*(-1)=9
sinx=(1-3)/4=-1/2        x=(-1)^(n+1)*π/6+πn, n∈Z
sinx=(1+3)/4=1          x=π/2+2πk,  k∈Z
3)5sin²x+3sinxcosx+4cos²x=3
5sin²x+3sinxcosx+4cos²x-3(sin²x+cos²x)=0
5sin²x+3sinxcosx+4cos²x-3sin²x-3cos²x=0 однородное, разделим на cos²x
2sin²x+3sinxcosx+cos²x=0 |  :  cos²x
2tg²x+3tgx+1=0
D=9-4*2*1=1
tgx=(-3-1)/4=-1    x=-π/4+πn,  n∈Z
tgx=(-3+1)/4=-1/2    x=-arctg1/2+πk,  k∈Z

Answer 2

1. To solve the equation 49^x - 6 - 7^x - 7 = 0, we can rearrange the terms to get:

49^x - 7^x = 13

Then, we can notice that 49^x can be expressed as (7^2)^x = 7^(2x), and 7^x can be factored out as a common factor:

7^x (7^x - 1) = 13

Solving for 7^x - 1, we get:

7^x - 1 = 13 / 7^x

Multiplying both sides by 7^x, we get:

7^(2x) - 7^x - 13 = 0

This is a quadratic equation in terms of 7^x. We can use the quadratic formula to solve for 7^x:

7^x = (1 ± √(1 + 4131)) / 2

Simplifying, we get:

7^x = (1 ± √53) / 2

Taking the logarithm base 7 of both sides, we get:

x = log_7[(1 ± √53) / 2]

So the solutions for the equation are:

x = log_7[(1 + √53) / 2] or x = log_7[(1 - √53) / 2]

2. To solve the equation cos(2x) + sin(x) = 0, we can use the identity:

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

Substituting this into the equation, we get:

1 - 2sin^2(x) + sin(x) = 0

Rearranging, we get:

2sin^2(x) - sin(x) + 1 = 0

This is a quadratic equation in terms of sin(x). We can use the quadratic formula to solve for sin(x):

sin(x) = [1 ± √(1 - 421)] / (2*2)

Simplifying, we get:

sin(x) = (1 ± i√7) / 4

Since sine is a real-valued function, the only solution is:

sin(x) = (1 - i√7) / 4

To find the corresponding value of x, we can use the inverse sine function:

x = arcsin[(1 - i√7) / 4]

Note that this solution involves complex numbers.

3. To solve the equation 5sin^2(x) = 3sin(x)cos(x) + 4cos^2(x) = 3, we can use the identity:

sin(2x) = 2sin(x)cos(x)

Substituting this into the equation, we get:

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

Simplifying, we get:

3sin(x)cos(x) - 3sin^2(x) + 3cos^2(x) = 0

Using the identity sin^2(x) + cos^2(x) = 1, we can simplify further:

3sin(x)cos(x) - 3 + 3cos^2(x) = 0

Dividing both sides by 3cos(x), we get:

sin(x) - 1/cos(x) = 1

Using the identity 1/cos(x) = sec(x), we can rewrite the equation as:

sin(x) - sec