Question

Sin 5x + sin 13x = 0 5 sin x -cos 2x +3 =0

Answer 1

Ответ:

Объяснение:

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Answer 2

To solve the trigonometric equations, let's solve them one by one:

1. Sin(5x) + Sin(13x) = 0

To solve this equation, we'll use the sum-to-product trigonometric identity:

Sin(A) + Sin(B) = 2 * Sin((A + B) / 2) * Cos((A - B) / 2)

So the equation becomes:

2 * Sin((5x + 13x) / 2) * Cos((5x - 13x) / 2) = 0

Now, simplify the equation:

2 * Sin(9x / 2) * Cos(-4x / 2) = 0

Since Sin(0) = 0, the equation reduces to:

Sin(9x / 2) = 0

To find the solutions for x, we need to find all angles θ such that Sin(θ) = 0. These angles are of the form θ = n * π, where n is an integer.

So, we have:

9x / 2 = n * π

Solving for x:

x = (2 * n * π) / 9

where n is an integer.

2. 5 * Sin(x) - Cos(2x) + 3 = 0

To solve this equation, we'll use some trigonometric identities:

Cos(2x) = 2 * Cos^2(x) - 1

Substitute this into the original equation:

5 * Sin(x) - (2 * Cos^2(x) - 1) + 3 = 0

Rearrange the equation:

2 * Cos^2(x) - 5 * Sin(x) + 4 = 0

Now, let u = Cos(x), the equation becomes:

2 * u^2 - 5 * Sin(x) + 4 = 0

To solve this quadratic equation for u, we can use the quadratic formula:

u = [ -b ± √(b^2 - 4ac) ] / 2a

where a = 2, b = -5, and c = 4.

Calculate the discriminant (D):

D = b^2 - 4ac D = (-5)^2 - 4 * 2 * 4 D = 25 - 32 D = -7

Since the discriminant is negative, there are no real solutions for u (Cos(x)).

Thus, the equation 5 * Sin(x) - Cos(2x) + 3 = 0 has no real solutions for x.