2sin^2×3x+5cos×3x+1=0

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вопрос

To solve the trigonometric equation 2sin^2(3x) + 5cos(3x) + 1 = 0, we can make use of the trigonometric identity sin^2(x) + cos^2(x) = 1.

Here's a step-by-step solution:

Rewrite 2sin^2(3x) as 2(1 - cos^2(3x)) using the identity sin^2(x) = 1 - cos^2(x):

2(1 - cos^2(3x)) + 5cos(3x) + 1 =
2 - 2cos^2(3x) + 5cos(3x) + 1 = 0

Rearrange the terms and rewrite the equation in terms of cos(3x):

-2cos^2(3x) + 5cos(3x) + 3 = 0

Let y = cos(3x), then the equation becomes a quadratic equation in terms of y:

-2y^2 + 5y + 3 = 0

Solve the quadratic equation for y by factoring or using the quadratic formula:

-2y^2 + 5y + 3 =
(-2y + 3)(y + 1) = 0

Hence, y = 3/2 or y = -1

Recall that y = cos(3x), so we have two cases to consider:

Case 1: cos(3x) = 3/
However, since the range of the cosine function is [-1, 1], there are no real solutions for this case.

Case 2: cos(3x) = -
This implies that 3x = π, adding 2nπ where n is an integer.

So the general solution for the equation 2sin^2(3x) + 5cos(3x) + 1 = 0 is
x = π/3 + (2nπ) / 3, where n is an integer.