Question

Sqrt(2)sin(x+pi/4)+cos2x=sinx-1

Answer 1

Решение в приложении.

Answer 2

Let's solve the given equation step by step.

The equation is:

√2sin(x + π/4) + cos(2x) = sin(x) - 1

First, let's simplify the equation:

√2sin(x + π/4) + cos(2x) = sin(x) - 1

Using the double-angle formula for cosine, we can write cos(2x) as 2cos²(x) - 1:

√2sin(x + π/4) + 2cos²(x) - 1 = sin(x) - 1

Rearranging the terms, we have:

√2sin(x + π/4) - sin(x) + 2cos²(x) = 0

Next, let's use the angle addition formula for sine:

√2(sin(x)cos(π/4) + cos(x)sin(π/4)) - sin(x) + 2cos²(x) = 0

Simplifying further, we get:

√2(sin(x)(√2/2) + cos(x)(√2/2)) - sin(x) + 2cos²(x) = 0

Simplifying the coefficients, we have:

(sin(x) + cos(x))√2 - sin(x) + 2cos²(x) = 0

Expanding the brackets, we get:

√2sin(x) + √2cos(x) - sin(x) + 2cos²(x) = 0

Combining like terms, we obtain:

(√2 - 1)sin(x) + (√2cos(x) + 2cos²(x)) = 0

Now, let's factor out sin(x) from the first term and cos(x) from the second term:

sin(x)(√2 - 1) + cos(x)(√2 + 2cos(x)) = 0

At this point, we have two possibilities:

1. sin(x) = 0
2. (√2 - 1) + (√2 + 2cos(x)) = 0

Case 1: sin(x) = 0

If sin(x) = 0, it means x is a multiple of π. Therefore, the solutions in this case are:

x = nπ, where n is an integer.

Case 2: (√2 - 1) + (√2 + 2cos(x)) = 0

Rearranging the terms, we get:

√2 + 2cos(x) = 1 - √2

Simplifying further:

2cos(x) = 1 - 2√2

cos(x) = (1 - 2√2) / 2

cos(x) = 0.29289 (approx.)

Using inverse cosine (arccos), we can find the values of x:

x = arccos(0.29289) + 2πk or x = -arccos(0.29289) + 2πk, where k is an integer.

Please note that the above values are approximate and may need to be rounded to a certain number of decimal places depending on the desired level of precision.

Therefore, the solutions to the given equation are:

x = nπ or x = arccos(0.29289) + 2πk or x = -arccos(0.29289) + 2πk, where n and k are integers.