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Solve for y in terms of x in the given trigonometric equation:

sin(3x) * 2cos(2x)sin(3x) = sin²(y) * sin²(x)
a) y = sin^(-1)(3x)
b) y = sin^(-1)(2x)
c) y = sin^(-1)(x)
d) y = cos^(-1)(x)

1 Answer

3 votes

Final answer:

The correct answer to the trigonometric equation is not provided among the choices. Solving the equation by simplification suggests that y is directly proportional to x. Thus, y is ±3x, which is not in the form of any arcsine function provided.

Step-by-step explanation:

Let's solve the trigonometric equation for y in terms of x. The given equation is:

sin(3x) × 2cos(2x)sin(3x) = sin²(y) × sin²(x)

First, we can simplify the left side using the trigonometric identity sin(2α) = 2sin(α)cos(α), applied to 3x:

sin(3x) × 2cos(2x)sin(3x) = sin(3x) × sin(6x)

Now, let's assume that y and x are in the range where the sine function is bijective and has an inverse (for example, [-π/2, π/2]), so that we can apply the inverse sine function. Note that the equation now simplifies to:

sin²(3x) = sin²(y) × sin²(x)

Since we are looking for the solution where sin(y) is equal to sin(3x) or -sin(3x), we get:

y = sin^(-1)(sin(3x))
But since sin(y) could be negative, we also consider:
y = -sin^(-1)(sin(3x))

However, given the principal value range for sin^(-1), the correct representation for y would be:

y = ±3x

Now, since we're solving for y in terms of x and not sin(3x), the provided options are comparing the value of y to known angles. Thus, the correct answer is none of the provided multiple-choice answers, because y is not dependent on the arcsine of a value, but rather directly related to x itself.

User Alvaro Cavalcanti
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