3sinx+5sin2x=0

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вопрос

To solve the equation 3sin(x) + 5sin(2x) = 0, we can first use the double angle formula for sine, which states that sin(2x) = 2sin(x)cos(x).

Substitute sin(2x) = 2sin(x)cos(x) into the equation:

3sin(x) + 5(2sin(x)cos(x)) = 0
3sin(x) + 10sin(x)cos(x) = 0
sin(x)(3 + 10cos(x)) = 0

Now we have two possibilities:

For sin(x) = 0, the solution is x = kπ, where k is an integer.

For 3 + 10cos(x) = 0, we can solve for cos(x):

10cos(x) = -3
cos(x) = -3/10

Since cos(x) = -3/10, we can use inverse trigonometric functions to find the solutions for x. The solutions for cos(x) = -3/10 lie in the second and third quadrants.

Therefore, x = π + arccos(-3/10) and x = 2π - arccos(-3/10). These are the solutions for the equation 3sin(x) + 5sin(2x) = 0.