Question

(sinx-cosx)^2=0,5-sinx cosx помогите

Answer 1

Sin(x)^2 - 2sin(x)cos(x) + cos(x)^2 = 0.5 - sin(x)cos(x)
1 - 2sin(x)cos(x) = 0.5 - sin(x)cos(x)
sin(x)cos(x) = 0.5 sin(x)^2 + 0.5 cos(x)^2
0.5sin(x)^2 + 0.5cos(x)^2 - sin(x)cos(x) = 0
cos(x) не равен нулю, ведь тогда и sin(x) = 0, а этого быть не может. Разделим на cos(x)^2
0.5 tg(x)^2 -tg(x) + 0.5 = 0
tg(x)^2 - 2tg(x) + 1 = 0
Сделаем замену: t = tg(x)
t^2 - 2t + 1 = 0
D = 4 - 4 = 0
t = 1
tg(x) = 1
x = p/4 + n*p

Answer 2

To solve this equation, we can start by expanding the left-hand side using the identity:

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

Then, we can simplify the right-hand side using the identity:

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

Now we have:

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

Expanding both sides:

2 sin^2 x - 4 sin x cos x + 2 cos^2 x = sin^2 x - 2 sin x cos x + cos^2 x

Simplifying:

sin^2 x - 2 sin x cos x + cos^2 x = 0

This can be factored as:

(sin x - cos x)^2 = 0

Taking the square root of both sides:

sin x - cos x = 0

or

sin x = cos x

To solve for x, we can take the inverse sine of both sides:

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

x = π/4 + nπ, where n is an integer.

Therefore, the solutions to the original equation are all values of x that satisfy x = π/4 + nπ, where n is an integer.