Question

Sinx+cosx=1+0,5sin2x

Answer 1

Sinx+cosx=1
1+0,5sin2x=1
0,5sin2x=0
sin2x=0
2x=kП, kєZ
x=kП/2

Answer 2

To solve the equation sin(x) + cos(x) = 1 + 0.5sin(2x), we'll first simplify the right-hand side by using the double-angle identity for sine (sin(2x) = 2sin(x)cos(x)):

sin(x) + cos(x) = 1 + 0.5(2sin(x)cos(x))

Now, let's move all terms to one side of the equation:

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

Next, let's use the identity for sin(x) - cos(x) to simplify the left-hand side:

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

Now, recall the Pythagorean identity: sin^2(x) + cos^2(x) = 1

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

We can rewrite the equation as:

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

Now, let's solve for sin(x)cos(x):

2sin(x)cos(x) = 1

sin(x)cos(x) = 1/2

Now, we know that sin(2x) = 2sin(x)cos(x), so we can replace sin(x)cos(x) with 1/2 in the original equation:

sin(2x) = 2(1/2)

sin(2x) = 1

Now, we have a simple equation for sin(2x). The solutions for sin(2x) = 1 are:

1. When 2x = π/2 + 2πk, where k is an integer.
2. When 2x = 3π/2 + 2πk, where k is an integer.

Now, let's find the solutions for x:

1. When 2x = π/2 + 2πk: x = (π/2 + 2πk)/2 x = π/4 + πk

2. When 2x = 3π/2 + 2πk: x = (3π/2 + 2πk)/2 x = 3π/4 + πk

So, the solutions for the equation sin(x) + cos(x) = 1 + 0.5sin(2x) are:

x = π/4 + πk and x = 3π/4 + πk, where k is an integer.