Cos2x-3sinxcosx+sin2x=0

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вопрос

To solve this trigonometric equation, we first want to simplify and rewrite it in terms of either sine or cosine.

Given: Cos(2x) - 3sin(x)cos(x) + sin(2x) = 0

Using trigonometric identities:
Cos(2x) = 1 - 2sin^2(x)
Sin(2x) = 2sin(x)cos(x)

We can rewrite the equation as:
1 - 2sin^2(x) - 3sin(x)cos(x) + 2sin(x)cos(x) = 0

Now, we can combine like terms and simplify:
1 - 2sin^2(x) - sin(x)cos(x) = 0

Next, we can use the double angle formula for sine and cosine functions:
Sin(2x) = 2sin(x)cos(x)

Substitute this back into the equation:
1 - 2sin^2(x) - sin(2x) = 0

Now, let u = sin(x), then we have:
1 - 2u^2 - 2u = 0

Rearrange the equation:
2u^2 + 2u - 1 = 0

Factor the quadratic equation:
(2u - 1)(u + 1) = 0

Now, solve for u:
2u - 1 = 0
2u = 1
u = 1/2

u + 1 = 0
u = -1

Now substitute back u = sin(x):
sin(x) = 1/2
x = π/6 + 2nπ , 11π/6 + 2nπ, n ∈ Z

sin(x) = -1
x = 3π/2 + 2nπ, n ∈ Z

Therefore, the solutions to the trigonometric equation are x = π/6, 11π/6, 3π/2 and any integer times 2π.