Question

Sinxcosx+3sin2x-cos2x=0

Answer 1

Ответ:

Ответ на рисунке, 5 фотографий, решение есть

Answer 2

Ответ: во вложении Пошаговое объяснение:

Answer 3

We can start by simplifying each term using trigonometric identities:

sinxcosx = 1/2 sin(2x) 3sin2x = 3(2sinx cosx) = 3sinx cosx + 3sinx cosx = 3/2 sin(2x) cos2x = cos^2(x) - sin^2(x) = 1 - 2sin^2(x)

Substituting these expressions into the original equation, we get:

1/2 sin(2x) + 3/2 sin(2x) - (1 - 2sin^2(x)) = 0

Simplifying further:

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

This is now a quadratic equation in sin(x), which we can solve using the quadratic formula:

sin(x) = [-b ± sqrt(b^2 - 4ac)] / 2a

where a = 2, b = 1, and c = -1. Substituting these values, we get:

sin(x) = [-1 ± sqrt(1 + 8)] / 4

sin(x) = [-1 ± sqrt(9)] / 4

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

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

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

x = -π/6 or x = π/6

Therefore, the solutions to the original equation are:

x = -π/6 + kπ or x = π/6 + kπ, where k is any integer.